course Mth 271 I apologize for this assignment being late. It was a bad week and I didn't get a chance to work on it till later. o̾ϸƯassignment #012
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14:57:29 `qNote that there are 12 questions in this assignment. `q001. When we form the composite of two functions, we first apply one function to the variable, then we apply the other function to the result. We can for example first apply the function z = t^2 to the variable t, then we can apply the function y = e^z to our result. If we apply the functions as specified to the values t = -2, -1, -.5, 0, .5, 1, 2, what y values to we get?
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RESPONSE --> For t=-2: z=(-2)^2 = 4 ; y= e^4= 54.5982 For t=-1: z=(-1)^2 = 1 ; y= e^1= 2.7183 For t=-0.5: z=(-0.5)^2 = 0.25 ; y= e^0.25 = 1.2840 For t=1 z=1^2 = 1 ; y= e^1 = 2.7183 For t=2: z=2^2 = 4 ; y= e^4 = 54.5982 confidence assessment: 3
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14:58:14 If t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx.. If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1. If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx..
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RESPONSE --> The question did not have t=0 as a variable to determine. self critique assessment: 2
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15:04:11 `q002. If we evaluate the function y = e^(t^2) at t = -2, -1, -.5, 0, .5, 1, 2, what y values do we get?
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RESPONSE --> For t=-2: y= e^[(-2)^2] = 54.60 For t=-1: y= e^[(-1)^2] = 2.72 For t=-0.5: y= e^[(-0.5)^2] = 1.28 For t=0: y= e^(0^2) = 1 For t=0.5 y= e^(0.5^2) = 1.28 For t=1: y= e^(1^2) = 2.72 For t=2: y= e^(2^2) = 54.60 confidence assessment: 3
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15:04:24 If t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx. If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx. If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx. If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1. If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx. If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx. If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx.
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RESPONSE --> ok self critique assessment: 3
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15:06:29 `q003. We see from the preceding two examples that that the function y = e^(t^2) results from the 'chain' of simple functions z = t^2 and y = e^z. What would be the 'chain' of simple functions for the function y = cos ( ln(x) )?
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RESPONSE --> z = ln(x) and y = cos(z) confidence assessment: 3
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15:07:06 The first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z). Thus we have y = cos(z) = cos( ln(x) ). We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x).
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RESPONSE --> ok self critique assessment: 3
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15:08:30 `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?
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RESPONSE --> z = t^2 and y = ( ln(t) ) ^z confidence assessment: 2
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15:09:15 The first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2. Thus we have y = z^2 = (ln(t))^2. We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t).
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RESPONSE --> I had the variables backwards. I do understand this concept. self critique assessment: 2
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15:11:07 `q005. What would be the chain of functions for y = ln ( cos(x) )?
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RESPONSE --> z = cos(x) and y = ln(z) confidence assessment: 2
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15:11:21 The first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z). Thus we have y = ln(z) = ln(cos(x)). This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x).
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RESPONSE --> ok self critique assessment: 3
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15:20:30 `q006. The rule for the derivative of a chain of functions is as follows: The derivative of the function y = f ( g ( x) ) is y ' = g' ( x ) * f ' ( g ( x) ). For example if y = cos ( x^2 ) then we see that the f function is f(z) = cos(z) and the g function is the x^2 function, so that f ( g ( x) ) = f ( x^2 ) = cos ( x^2 ) . By the chain rule the derivative of this function will be (cos(x^2)) ' = g ' ( x) * f ' ( g ( x) ) . g(x) = x^2 so g'(x) = 2 x. f ( z ) = cos ( z) so f ' ( z ) = - sin( z ), so f ' ( g ( x ) ) = - sin ( g ( x ) ) = - sin ( x^2). Thus we obtain the derivative (cos(x^2)) ' = g ' ( x ) * f ' ( g ( x ) ) = 2 x * ( - sin ( x^2 ) ) = - 2 x sin ( x^2). Apply the rule to find the derivative of y = sin ( ln ( x ) ) .
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RESPONSE --> The derivative of y = sin(ln(x)) is y`= cos(x) * 1 / ( sin(x) ) confidence assessment: 2
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15:22:08 We see that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ). Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x. Since f(z) = sin(z) we have f ' (z) = cos(z). Thus the derivative of y = sin( ln (x) ) is y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) = 1 / x * cos( ln(x) ). Note how the derivative of the 'inner function' g(x) = ln(x) appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the sine function.
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RESPONSE --> I got the f(x) and g(x) mixed up but I would have gotten the right answer had I not done that. self critique assessment: 2
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15:26:42 `q007. Find the derivative of y = ln ( 5 x^7 ) .
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RESPONSE --> The derivative of y = ln( 5x^7 ) is y` = 45x^6 * 1 / ( 5x^7) with f(x) = ln(x) and f`(x) = 1/x and g(x) = 5x^7 and g`(x) = 45x^6 confidence assessment: 2
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15:27:34 For this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus f ' (z) = 1 / z and g ' (x) = 35 x^6. We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7). So the derivative of y = ln( 5 x^7) is y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ]. Note again how the derivative of the 'inner function' g(x) = 5 x^7 appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the reciprocal or 1 / z function.
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RESPONSE --> I did my math wrong on the 5*7 part but other than that I had the problem right. self critique assessment: 2
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15:30:35 `q008. Find the derivative of y = e ^ ( t ^ 2 ).
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RESPONSE --> The derivative of y = e^(t^2) is y`= 2t * e^(t^2) with f(x) = e^z and f`(x) = e^z and g(x) = t^2 and g`(x) = 2t confidence assessment: 2
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15:30:59 This function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t. Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2). Note once more how the derivative of the 'inner function' g(t) = t^2 appears 'out in front' of the derivative of the 'outer' function.
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RESPONSE --> ok self critique assessment: 3
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15:33:54 `q009. Find the derivative of y = cos ( e^t ).
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RESPONSE --> The derivative of y = cos( e^t ) is y` = e^t * -sin( e^t ) with f(x) = cos(x) and f`(x) = -sin(x) and g(x) = e^t and g`(x) = e^t confidence assessment: 2
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15:34:27 We have the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t. Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t). Note how the 'inner function' is unchanged, as it has been in previous examples, and how its derivative appears in front of the derivative of the 'outer' function.
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RESPONSE --> I did not completely factor my answer but it was right. self critique assessment: 2
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15:38:39 `q010. Find the derivative of y = ( ln ( t ) ) ^ 9.
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RESPONSE --> The derivative of y = ( ln (t) ) ^ 9 is y` = 9t^8 * 1/ (t^9) with f(x) = ln(t) and f`(x) = 1/t and g(x) = t^9 and g`(x) = 9t^8 confidence assessment: 2
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15:40:38 We have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8.
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RESPONSE --> I got the f(x) and the g(x) terms confused again. I will be sure to pay extra attention to the problems that I seem to keep getting backwards. self critique assessment: 2
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15:44:53 `q011. Find the derivative of y = sin^4 ( x ).
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RESPONSE --> The derivative of y = sin ^4 (x) is y`= cos(x) * 4(cos(x))^3 ??? this is a tricky one confidence assessment: 1
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15:47:41 The composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power. We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x).
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RESPONSE --> I did not realize that f(x) could still be z^4 with the sin^4. But it is worked the same way as the other problems and I get it now. self critique assessment: 2
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15:51:06 `q012. Find the derivative of y = cos ( 3x ).
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RESPONSE --> The derivative of y = cos(3x) is y`= -sin(x) * 3(cos(x)) = -3sin(x)cos(x) with f(x) = 3x and f`(x) = 3 and g(x) = cos(x) and g`(x) = -sin(x) confidence assessment: 2
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15:52:59 This is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).
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RESPONSE --> Agian I got my f(x) and g(x) mixed up but I did this problem again with the right f(x) and g(x) and got the right answer. self critique assessment: 2
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N^Nٽnԉ̇jqֲT assignment #012 012. `query 12 Applied Calculus I 10-19-2007
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15:54:59 Class Notes #13 Explain how we obtain algebraically, starting from the difference quotient, the expression for the derivative of the y = x^2 function.
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RESPONSE --> The derivative of y = x^2 is obtained algebraically by: ( f(x + `dx) - f(x) )/ `dx which gives y`= 2x for the derivative of y = x^2 confidence assessment: 2
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15:55:24 The difference quotient is [ f(x + `dx ) - f(x) ] / `dx. In this case we get [ (x+`dx)^2 - x^2 ] / `dx = [ x^2 + 2 x `dx + `dx^2 - x^2 ] / `dx = [ 2 x `dx - `dx^2 ] / `dx = 2 x - `dx. Taking the limit as `dx -> 0 this gives us just 2 x. y ' = 2 x is the derivative of y = x^2. **
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RESPONSE --> ok self critique assessment: 3
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15:58:03 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> The derivative of y = x^n is obtained using the binomial formula by: (x + `dx)^n which gives y= nx^(n-1) as the derivative of y = x^n confidence assessment: 2
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15:58:50 The key is that (x + `dx)^n = x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. When we form the difference quotient the numerator is therefore f(x+`dx) - f(x) = (x + `dx)^n - x^n = (x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) - x^n = n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. The difference quotient therefore becomes (n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) / `dx = n x^(n-1) + n (n-1) / 2 * x^(n-2) `dx + ... +`dx^(n-1). After the first term n x^(n-1) every term has some positive power of `dx as a factor. Therefore as `dx -> 0 these terms all disappear and the limiting result is n x^(n-1). **
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RESPONSE --> ok self critique assessment: 3
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16:01:55 **** Explain how the derivative of y = x^3 is used in finding the equation of a tangent line to that graph.
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RESPONSE --> The derivative of y = x^3 is determined by finding slope. This shows that the slope of the tangent line is the derivative of the equation. confidence assessment: 2
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16:02:04 The derivative is the slope of the tangent line. If we know the value of x at which we want to find the tangent line then we can find the coordinates of the point of tangency. We evaluate the derivative to find the slope of the tangent line. Know the point and the slope we use the point-slope form to get the equation of the tangent line. **
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RESPONSE --> ok self critique assessment: 3
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16:02:42 You can use any two points on the graph to estimate the slope. The slope of a straight line is the same no matter what two points you use. Of course estimates can vary; the common approach of moving over 1 unit and seeing how many units you go up is a good method when the scale of the graph makes it possible to accurately estimate the distances involved. The rise and the run should be big enough that you can obtain good estimates. One person's estimate: my estimate is -1/3. I obtained this by seeing that for every 3 units of run, the tangent line fell by one unit. Therefore rise/run = -1/3. **
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RESPONSE --> this part skipped to the answer before I could answer the question but I did get m = -1/3 as an estimate for the slope self critique assessment: 2
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16:07:53 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> This is one of the problems that took me a while to do but I got y`=3t^2 + 2t confidence assessment: 2
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16:08:13 f(t+`dt) = (t+'dt)^3+(t+'dt)^2. f(t) = t^3 + t^2. So [f(t+`dt) - f(t) ] / `dt = [ (t+'dt)^3+(t+'dt)^2 - (t^3 + t^2) ] / `dt. Expanding the square and the cube we get [t^3+3t^2'dt+3t('dt)^2+'dt^3]+[t^2+2t'dt+'dt^2] - (t^3 - t^2) } / `dt. } We have t^3 - t^3 and t^2 - t^2 in the numerator, so these terms subtract out, leaving [3t^2'dt+3t('dt)^2+'dt^3+2t'dt+'dt^2] / `dt. Dividing thru by `dt you are left with 3t^2+3t('dt)+'dt^2+2t+'dt. As `dt -> 0 you are left with just 3 t^2 + 2 t. **
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RESPONSE --> ok self critique assessment: 3
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16:14:37 2.1.32 tan line to y = x^2+2x+1 at (-3,4) What is the equation of your tangent line and how did you get it?
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RESPONSE --> The equation of the tangent line was figured by: f(x) = x^2 + 2x + 1 at point (-3,4) substituting -3 for x => -3^2 + 2(3) + 1 = = 9 + -6 + 1 = 4 this was to check and make sure that the point was right. Then I put -3 in for x on the derivative of the equation which was f `(x)= 2x + x to get the slope of the line. f `(x)= 2(-3) + 2 = -4 Then I used the point slope form to find the equation: y - 4 = -4(x - (-3)) => y= -4x - 8 confidence assessment: 2
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16:15:01 STUDENT SOLUTION WITH INSTRUCTOR COMMENT: f ' (x)=2x+2. I got this using the method of finding the derivative that we learned in the modeling projects The equation of the tangent line is 2x+2. I obtained this equation by using the differential equation. the slope is -4...i got it by plugging the given x value into the equation of the tan line. INSTRUCTOR COMMENT: If slope is -4 then the tangent line can't be y = 2x + 2--that line has slope 2. y = 2x + 2 is the derivative function. You evaluate it to find the slope of the tangent line at the given point. You have correctly found that the derivative is -4. Now the graph point is (-3,4) and the slope is -4. You need to find the line with those properties--just use point-slope form. You get y - 4 = -4(x - -3) or y = -4 x - 8. **
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RESPONSE --> ok self critique assessment: 3
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16:17:28 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown) At what points is the function differentiable, and why?
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RESPONSE --> #52 was not assigned so it will take me a litte while longer to answer this question. x^2-4 divides out to be (x - 2) (x + 2) so the points that the function is differentiable is x = + or - 2 confidence assessment: 2
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16:17:40 10-19-2007 16:17:40 At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist. The reason the derivative doesn't exist for this function this is that the function isn't even defined at x = +- 2. So there if x = +- 2 there is no f(2) to use when defining the derivative as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. f(2+`dx) is fine, but f(2) just does not exist as a real number. **
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NOTES -------> ok
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16:17:42 At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist. The reason the derivative doesn't exist for this function this is that the function isn't even defined at x = +- 2. So there if x = +- 2 there is no f(2) to use when defining the derivative as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. f(2+`dx) is fine, but f(2) just does not exist as a real number. **
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RESPONSE --> self critique assessment: 3
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16:18:43 **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown)
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RESPONSE --> (-infinity , -2) , (-2,2) , (2, infinity) confidence assessment: 2
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16:18:59 The derivative is defined on(-infinity,-2)u(-2,2)u(2,infinity). The reason the derivative doesn't exist at x = +-2 is that the function isn't even defined at x = +- 2. The derivative at 2, for example, is defined as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. If f(2) is not defined then this expression is not defined. The derivative therefore does not exist. At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist.**
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RESPONSE --> ok self critique assessment: 3
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16:23:01 If x is close to but not equal to 2, what makes you think that the function is differentiable at x?
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RESPONSE --> Becuase the function doen't reach x it has limits that do not involve x. confidence assessment: 2
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16:23:18 If x is close to 2 you have a nice smooth curve close to the corresponding point (x, f(x) ), so as long as `dx is small enough you can define the difference quotient and the limit will exist. **
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RESPONSE --> ok self critique assessment: 3
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16:23:50 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> If x=2 the function is not differentiable because there is no limit confidence assessment: 2
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16:24:04 GOOD ANSWER FROM STUDENT: if the function does not have limits at that point then it is not differentiableat at that point.
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RESPONSE --> ok self critique assessment: 3
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