Knowee
Questions
Features
Study Tools

Suppose that it is given to you that f′(x)=(x+5)(6−x)(x−9) Then the first relative extremum (from the left) for f(x) occurs at x= The function f(x) has a relative ? at this point. The second relative extremum (from the left) for f(x) occurs at x= The function f(x) has a relative ? at this point. The third relative extremum (from the left) for f(x) occurs at x= The function f(x) has a relative ? at this point. The first inflection point (from the left) for f(x) occurs at x= The second inflection point (from the left) for f(x) occurs at x=

Question

Suppose that it is given to you that f′(x)=(x+5)(6−x)(x−9)

Then the first relative extremum (from the left) for f(x) occurs at x=

The function f(x) has a relative ? at this point.

The second relative extremum (from the left) for f(x) occurs at x=

The function f(x) has a relative ? at this point.

The third relative extremum (from the left) for f(x) occurs at x=

The function f(x) has a relative ? at this point.

The first inflection point (from the left) for f(x) occurs at x=

The second inflection point (from the left) for f(x) occurs at x=

...expand
🧐 Not the exact question you are looking for?Go ask a question

Solution

The derivative of the function f(x) is given as f′(x)=(x+5)(6−x)(x−9). The critical points of the function, where the derivative is zero or undefined, are the solutions to the equation (x+5)(6−x)(x−9) = 0.

Solving this equation gives us x = -5, 6, and 9. These are the points where the function may have a relative extremum.

To determine whether these points are relative extrema and whether they are maxima or minima, we can use the first derivative test. This involves checking the sign of the derivative to the left and right of each critical point.

For x = -5, the derivative to the left is negative and to the right is positive. Therefore, the function has a relative minimum at x = -5.

For x = 6, the derivative to the left is positive and to the right is negative. Therefore, the function has a relative maximum at x = 6.

For x = 9, the derivative to the left is negative and to the right is positive. Therefore, the function has a relative minimum at x = 9.

The inflection points of the function are where the second derivative changes sign. The second derivative is the derivative of the first derivative, which can be found by applying the product and chain rules of differentiation to f′(x)=(x+5)(6−x)(x−9).

Without the explicit form of the second derivative, it's not possible to find the exact values of the inflection points. However, they will be located between the critical points, so we can say that the first inflection point will be between x = -5 and x = 6, and the second inflection point will be between x = 6 and x = 9.

This problem has been solved

Similar Questions

Consider the following function.f(x) = x2 + 3.5x − 6(a) Write the derivative formula.f '(x) = 2x+3.5 (b) Locate any relative extreme points. (If an answer does not exist, enter DNE.)relative maximum      (x, y) = relative minimum (x, y) =

Find all relative extrema of the function. Use the second derivative test where applicable. (If an answer does not exist, enter DNE.)f(x) = x4 − 8x3 + 1

Suppose that 4 ≤ f ′(x) ≤ 5 for all values of x. What are the minimum and maximum possible values of f(6) − f(3)? ≤ f(6) − f(3)

For the given function, find the requested relative extrema or extreme value (without a calculator).y = xe4x; relative extrema

Estimate the input value(s) where the function has a relative extreme point. Identify each relative extreme as a maximum or minimum, and indicate whether the derivative of the function at that point is zero or does not exist.input value maximumor minimum? derivativex = (smaller value) x = (larger value)

1/2

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.