Math 1500 Fall 2010 Final Exam Review Solutions

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1 Math 500 Fall 00 Final Eam Review Solutions. Verify that the function f() = 4 + on the interval [, 5] satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers c that satisfy the conclusion of the Mean Value Theorem. The function is continuous anywhere the is defined (which it is on this interval). It is also differentiable anywhere is defined, since the limit in the definition of the derivative eits. So the MVT tells us that there is some c in [, 5] so that f (c) = f(b) f(a) b a = f(5) f() 5 = =. To find c then we need to solve for f (c) = so c =. Solving for c we get c =. The graph of the derivative f of a function f is shown below. (a) On what intervals is f increasing or decreasing? The function increases when the derivative is positive and decreases when the derivative is negative so we look for when the graph is above or below the -ais. So this function increases on (,.) (0, ) (4., ) and decreases on (., 0) (, 4.). (b) At what values of does f have a local maimum or minimum? The First Derivative Test tells us that a function has a local maimum whenever the derivative change from positive to negative and a local minimum when the derivative changes from negative to positive. So, this would be places where the graph of the derivative crosses the -ais. So the local minimums are at = 0 and 4. and the local maimums are at =. and =. (c) On what intervals is f concave upward or concave downward?

2 Concavity is determined by the second derivative, so the function will be concave up wherever the second derivative is positive which means the first derivative is increasing. The function will be concave down wherever the second derivative is negative which means the first derivative is decreasing. So f is concave up on the region (.5,.5) (.5, ) while f is concave down on (,.5) (.5,.5).. Given the function f() = (a) What is the domain of f()? The domain is all so that is defined and not zero. So we need to solve > 0. Solving a quadratic inequality means finding where it is 0 and then applying a ±-chart. The chart for this function is below and so the domain is { R < or > } or all in (, ) (, ). (b) What are the intercepts of f()? From our answer to (a) we see that f(0) is undefined so there is no y-intercept. Similarly, for f() = 0 we would need the numerator to be 0 which only happens when = 0. So there is no -intercept. (c) Is f() even or odd or neither? f( ) = = ( ) = f() so the function is odd. 4. Given the function f() = ln ( ). (a) What is the domain of f()? First ln is only defined for positive. Second we need ( ) 0. So the domain is all real numbers greater than 0 and not equal to : { R > 0 and }. (b) Does f() have any horizontal asymptotes? We need to test lim f(). (Negative infinity does not make sense since the function is undefined for less than 0.) In this case we can apply L hospital s Rule since lim lim ( ) =. We get lim asymptote at y = 0. ln ( ) ( ) ln = and ( ) = 0. So there is a horizontal (c) Does f() have a vertical asymptote at =? If it has a vertical asymptote at = then lim f() must be ±. Since lim ln ( ) = 0, we can use L hospital s Rule to get lim f() ( ) ( ). Since the denominator

3 gets very, very close to 0, the limit must be ± (in one direction the denominator will be a small positive number, in the other it will be a small negative number) so f() does have a vertical asymptote at =. 5. Find the limits. e (a) lim +e cos Since lim e + e = 0 and lim cos = 0 we can use L hospital s Rule to get lim e + e cos e e sin. Now, lim e e = 0 and lim sin = 0 so we can apply L hospital s Rule again to get lim e e sin e + e 4 cos = 4 = by the Direct Substitution Property (since the numerator and denominator are no longer 0). (b) lim (e ) Since lim = and lim (e ) = 0 we can rewrite as lim (e e e ) and use L hospital s Rule to get: e =. (c) lim( + ) We first take a natural log of both sides of the function y = ( + ) to get ln y = So then lim ln y ln( + ) ln( + ). We can use L hospital s Rule since lim ln( + ) = 0 and lim = 0 so we get ln( + ) lim ln y. So since lim ln y =, lim ( + ) = e. + ( + ) = ln( + ). 6. Let f() = cos + on the interval 0 π. (a) Find the critical number(s) of the function. Critical numbers are where the derivative is zero or undefined. f () = sin() + critical numbers are where sin = so = π and π. and so the (b) Find the intervals on which f is increasing or decreasing.

4 By the Increasing-Decreasing Test, the function is increasing when the derivative is positive and decreasing when the derivative is negative. We make a ± chart between 0 and π with divisions at the critical numbers. Using our unit circle chart, we test f ( π 6 ) = + = +, f ( π ) = + =, and f (π) = 0 + = +. So the function is increasing on the intervals (0, π ) ( π, π) and decreasing on the interval ( π, π ). (c) Using both the first and second derivative test, find the local maimum and minimum values of f. By the First Derivative Test we know the local minimums occur at critical numbers where the derivative of the function changes from negative to positive. Looking at the chart in (b), this occurs at = π. Similarly there is a local ma at = π. For the Second Derivative Test, we must take the second derivative first. f () = cos() and so we plug the two critical numbers into f () to get f ( π ) = and f ( π ) =. The test then tells us = π is a local ma and = π is a local min. (d) Find the interval(s) where the function is concave upward and concave downward. A function is concave upward when the second derivative is positive and concave downward when the second derivative is negative. We already computed the second derivative f () = cos() and on the interval [0, π] cos() is positive in the first and fourth quardrant. So the function is concave up on the interval ( π, π ) and concave downward on the interval (0, π ) ( π, ). (e) Find the inflection point(s). The inflection points are where the concavity of the function changes so at = π and = π. 7. A cylindrical container, open at the top and of capacity 4π cubic inches is to be manufactured. If the cost of the material used for the bottom of the container is 6 cents per square inch, and the cost of the material used for the curved part is cents per square inch, find the dimension which will minimize the cost. (Hint: The bottom of the cylinder is a circle with area πr and the curved part is really a rectangle visualize cutting open a can and unfolding the curved part with height h and length the circumference of the bottom which is πr.) From the picture below we have labeled the radius r and height h. We know the volume of the cylinder is 4π = πr h. From the hint we know the cost of producing the bottom is 6 πr and the cost of producing the curved part is (πr h). Altogether we get : C = 6πr + πr h 4

5 And we can replace h with 4 r from the volume equation to get: C = 6πr + 4πr 4 r = 6πr + 96π r We take a derivative to see where this function is a minimum: C = πr 96π r = π (r 8r ) ( r ) 8 = π r. The critical number is when r = 8 or r =. We make a ± chart and test points C () = 7 and C () = 9 9 so by the First Derivative Test, r = is a minimum. To answer the question, we need to find h = 4 = 6. So the dimensions are r = and h = 6. 5