course Mth 173 I am still waiting to see if you have graded my second test (Test #1) 熽X蕩鳎G马榭秸鑯很欋疎邥朼ssignment #022
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08:59:46 Query problem 4.1.12 (3d edition 4.1.17) graph of e^x - 10x
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RESPONSE --> The graph has critical points of (.11183256, 0) and (3.5771521, 0). It has a minimum point of (2.302586, -13.02585). I found the derivative to be f'(x) = e^x - 10 I do not know how to use algebra to explain the graph.
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08:59:59 Explain the shape of the graph of e^x - 10x in terms of derivatives and algebra.
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RESPONSE --> In my last answer I said I did not know how to explain the graph.
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09:00:17 The derivative is y ' = e^x - 10, which is zero when e^x - 10 = 0, or e^x = 10. This occurs at x = ln(10), or approximately x = 2.30258. The second derivative is just y '' = e^x, which is always positive. The graph is therefore always concave up. Thus the 0 of the derivative at x = ln(10) implies a minimum of the function at (ln(10), 10 - 10 ln(10)), or approximately (2.30258, -13.0258). Since e^x -> 0 for large negative x, e^x - 10 x will be very close to -10 x as x becomes negative, so for negative x the graph is asymptotic to the line y = - 10 x.
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RESPONSE --> It slipped my mind that e^x = 10 can be solved for x with x = ln(10).
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09:00:29 Where is the first derivative positive, where is it negative and where is it zero, and how does the graph show this behavior?
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RESPONSE --> The first derivative, f'(x) = e^x - 10, is positive when x > ln(10), is negative when x < ln(10), and is zero when x = ln(10).
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09:00:41 ** The derivative e^x - 10 isn't always positive, though its graph does always have a positive slope. When this derivative is negative, as it is for x < 2.3 or so, the slope of the graph of e^x - 10 x will be negative so the graph will be decreasing. At the point where e^x - 10 becomes 0, indicating 0 slope for the graph of e^x - 10x, that graph reaches its minimum. For x > 2.3 (approx) the graph of e^x - 10 x will be increasing. **
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09:00:44 Where is the second derivative positive, where is in negative where is its zero, and how does the graph show this behavior?
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RESPONSE --> The second derivative, e^x, is always positive.
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09:00:56 ** The second derivative, e^x, is always positive. So the derivative is always increasing and that the function is always concave upward. **
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RESPONSE --> f(x) = a x e^(bx) f'(x) = a e^bx (1 + bx) f'(x) = 0 = -1/b Only critical point is -1 / b If that's the only critical point and the maximum is a x = 1/3 then -1 / b = 1/3 and b = -3 If b = -3 and f(1/3) = 1, then f(1/3) = a (1/3) e^(-3 * (1/3)) = 1; and a = 8.154845485
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09:00:57 ** The second derivative, e^x, is always positive. So the derivative is always increasing and that the function is always concave upward. **
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09:01:28 Query problem 4.1.29 (3d edition 4.1.26) a x e^(bx) What are the values of a and b such that f(1/3) = 1 and there is a local maximum at x = 1/3?
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RESPONSE --> f(x) = a x e^(bx) f'(x) = a e^bx (1 + bx) f'(x) = 0 = -1/b Only critical point is -1 / b If that's the only critical point and the maximum is a x = 1/3 then -1 / b = 1/3 and b = -3 If b = -3 and f(1/3) = 1, then f(1/3) = a (1/3) e^(-3 * (1/3)) = 1; and a = 8.154845485
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09:01:43 ** At a local maximum the derivative is zero so we have y'(1/3)=0. y ' (x) = a e^(bx) + a b x e^(bx) so if x = 1/3 we have ae^(1/3 b)+a b * 1/3 * e^(1/3b)=0. Factoring out ae^(1/3 * b) we get 1+1/3 b=0 which we easily solve for b to obtain b = -3. So now the function is y = a x e^(-3 x). We also know that f(1/3) = 1 so a * 1/3 e^(-3 * 1/3) = 1 or just a / 3 * e^-1 = 1, which is the same as a / ( 3 * e) = 1. We easily solve for a, obtaining a = 3 * e. So the function is now y = a x e^(-3x) = 3 * e * x e^(-3x). We can combine the e and the e^(-3x) to get y = 3 x e^(-3x+1). **
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09:06:57 Query problem 4.2.23 (3d edition 4.2.18) family x - k `sqrt(x), k >= 0. Explain how you showed that the local minimum of any such function is 1/4 of the way between its x-intercepts.
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RESPONSE --> The derivative is 1 - k/(2 'sqrt x) f'(x) = 0; x = k^2 / 4 f''(x) = 2 / k^2: This is greater than zero so there is a minimum at this critical point. x - kx^.5 = 0 By factoring we can find that x = 0 and x = k^2 I plotted k^2 / 4, x = 0, and x = k^2 and it is visible that k^2 / 4 is 1/4 of the way between the intercepts.
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09:07:08 After you find that the critical pt is k^2/4 y '' (k^2 /4) =2/k^2, which is greater than zero, so it is positive and there is a minimum at this crit pt for the zeros: x-kx^.5=0 x=0 or x=k^2 after factoring By plotting a no line, you can see that k^2 /4 is 1/4 of the way between 0 and k^2 ** Plotting gives you a good visual depiction but 1/4 of the way between 0 and anything is 1/4 * anything, so 1/4 of the way between 0 and k^2 is 1/4 k^2, which is what you found for the critical point. **
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09:09:20 What are the x-intercepts of f(x) = x - k `sqrt(x) and how did you find them?
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RESPONSE --> I first divided both sides by sqrt(x) to get 'sqrt(x) - k = 0. ( x / 'sqrt(x) = x * x^-1/2 = x^1/2 = 'sqrt(x)) Then I solved for x to find that x = k^2
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09:09:45 ** This problem requires the use of derivatives and other calculus-based tools (which do not include the graphing calculator). You have to find intercepts, critical points, concavity, etc.. We first find the zeros: x-intercept occurs when x - k `sqrt(x) = 0, which we solve by by dividing both sides by `sqrt(x) to get `sqrt(x) - k = 0, which we solve to get x = k^2. So the x intercept is at (k^2, 0). **
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09:14:07 04-21-2008 09:14:07 Where is(are) the critical point(s) of x - k `sqrt(x) and how did you find it(them)?
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NOTES -------> The derivative of x - k `sqrt(x) is 1 - k / (2 `sqrt(x)) I then set f'(x) = 0. 1 - k / (2 `sqrt(x)) = 0 I first multiplied both sides by 2'sqrt(x) to get 2'sqrt(x) - k = 0 2 'sqrt(x) = k 'sqrt(x) = k/2 x = k^2 / 4 This is the critical point.
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09:14:10 Where is(are) the critical point(s) of x - k `sqrt(x) and how did you find it(them)?
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09:14:37 ** We now find the critical point, where f ' (x) = 0: If f(x) = x - k `sqrt(x) then f ' (x) = 1 - k / (2 `sqrt(x)). f'(x) = 0 when 1 - k / (2 `sqrt(x)) = 0. Multiplying both sides by the denominator we get 2 `sqrt(x) - k = 0 so that x = k^2 / 4. The critical point occurs at x = k^2 / 4. If k > 0 the second derivative is easily found to be positive at this point, so at this point we have a minimum. The derivative is a large negative number near the origin, so the graph starts out steeply downward from the origin, levels off at x = k^2 / 4 where it reaches its minimum, then rises to meet the x axis at (k^2, 0). We note that the minumum occurs 4 times closer to the origin than the x intercept. **
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RESPONSE --> The derivative of x - k `sqrt(x) is 1 - k / (2 `sqrt(x)) I then set f'(x) = 0. 1 - k / (2 `sqrt(x)) = 0 I first multiplied both sides by 2'sqrt(x) to get 2'sqrt(x) - k = 0 2 'sqrt(x) = k 'sqrt(x) = k/2 x = k^2 / 4 This is the critical point.
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09:22:18 Query problem 4.2.37 (3d edition 4.2.31) U = b( a^2/x^2 - a/x). What are the intercepts and asymptotes of this function? At what points does the function have local maxima and minima? Describe the graph of the function.
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RESPONSE --> First find the dervative of U. U' = b(-2a^2 / x^3 + a / x^2) To find the intercepts I the derivative equal to zero. I solved for x. -2a^2 / x^3 + a / x^2 = 0; x = 2 a^2 / a = 2a After a frist-derivative test I found out that this point is a minimum and the function reaches its minimum at (2a, -b/4)
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09:22:21 ** We use the standard techniques to analyze the graphs: The derivative of U = b( a^2/x^2 - a/x) is dU/dx = b(-2a^2 / x^3 + a / x^2), which is 0 when -2a^2 / x^3 + a / x^2 = 0. This is easily solved: we get -2a^2 + a x = 0 so x = 2 a^2 / a = 2a. A first-or second-derivative test tells us that this point is a minimum. The function therefore reaches its minimum at (2a, -b/4). The parameter a determines the x coordinate of the minimum, and the parameter b independently determines the value of the minimum. Adjusting the constant a changes the x-coordinate of the minimum point, moving it further out along the x axis as a increases, without affecting the minimum. Adjusting b raises or lowers the minimum: larger positive b lowers the minimum, which for positive b is negative; and an opposite effect occurs for negative b. Therefore by adjusting two parameters a and b we can fit this curve to a wide variety of conditions. As x -> +-infinity, both denominators in the U function get very large, so that U -> 0. Thus the graph will have the x axis as an asymptote at both ends. As x -> 0, a^2 / x^2 will approach -infinity for positive a, while -a / x will approach + infinity. However, since near x=0 we see that x^2 is much smaller than x so its reciprocal will be much larger, the a^2 / x^2 term will dominate and the vertical asymptote will therefore be the positive y axis. As x gets large, x^2 will increasingly dominate x so the -a/x term will become larger than the a^2 / x^2 term, and if a is positive the graph must become negative and stay that way. Note that a graph of a smooth continuous function that becomes and stays negative while eventually approaching 0 is bound to have a minimum. **
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09:22:39 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> A lot of new stuff to learn, not too bad though.
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09:22:43 I did have some problems with the exercises, but I think I just need to practice more, especially on asymptotes and x and y intercepts
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