course Mth 173 4/9 3 Question: `qQuery problem 4.1.12 (3d edition 4.1.17) graph of e^x - 10xExplain the shape of the graph of e^x - 10x in terms of derivatives and algebra.
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The second derivative gives you the fact that the graph is positive, thus concave up with a min at y = -13.02. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qWhere is the first derivative positive, where is it negative and where is it zero, and how does the graph show this behavior? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The minimum is at x = 2.3 horizontally so the graph will be negative to the left of the min and positive to the right of the min. It will be zero at x = 2.3. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qWhere is the second derivative positive, where is in negative where is its zero, and how does the graph show this behavior? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = e^x – 10x F’ = e^x – 10 F’’ = e^x It is never negative. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The second derivative, e^x, is always positive. So the derivative is always increasing and that the function is always concave upward. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): positive and concave upward* ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 4.1.29 (3d edition 4.1.26) a x e^(bx) What are the values of a and b such that f(1/3) = 1 and there is a local maximum at x = 1/3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t quite understand, so f(1/3) = 1 and f(1/9) = the max??? confidence rating #$&* 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand the derivative rules used, but the simplification is a mystery to me.
I could try to explain the given solution further, but I'm
afraid in this case that would further complicate the issue.Write down, on paper, the equation
a e^(1/3 b) + a b * 1/3 * e^(1/3b) = 0
and solve it for b.
If you didn't get b = -3, then answer these questions:
What do you get when you divide both sides by a?
What do you get when you then divide both sides by
e^(1/3 b)?What is the solution to the resulting equation?
Then take another look at the given solution. If
you don't understand it after working it out yourself, submit an outline of what you did and I'll be happy to comment.
Then do the same with the equation
a * 1/3 e^(-3 * 1/3) = 1
Solve the equation for a. Note that e^(-3 * 1/3) =
e^(-1) = 1 / e. ------------------------------------------------ Self-critique rating #$&* 1 ********************************************* Question: `qQuery problem 4.2.23 (3d edition 4.2.18) family x - k `sqrt(x), k >= 0. Explain how you showed that the local minimum of any such function is 1/4 of the way between its x-intercepts. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To be honest, beforehand I wasn’t aware of this but all you would need to do is put the function over 4. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `a ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): or all over 4. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qWhat are the x-intercepts of f(x) = x - k `sqrt(x) and how did you find them? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: solve for x to get x = k^2. There is only one x-int and it is at constant^2. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qWhere is(are) the critical point(s) of x - k `sqrt(x) and how did you find it(them)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Critical pts are where the derivative equals 0. And that is 1 – k/2(sqrt(x)) = 0 0 = (k^2)/4 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 4.2.37 (3d edition 4.2.31) U = b( a^2/x^2 - a/x). What are the intercepts and asymptotes of this function? At what points does the function have local maxima and minima? Describe the graph of the function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: U = b( a^2/x^2 - a/x) U’ = b ((-2a^2) / x^3 + (a) / x^2) b ((-2a^2) / x^3 + (a) / x^2) = 0 x = 2(a) and its concave up, a minimum asymptotes will be the x and y axis confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): *x and y (positive) axis for vertical and horizontal asymptotes. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): N/A ------------------------------------------------ Self-critique rating #$&*N/A
&#Your work looks good. See my notes. Let me know if you have any questions. &#