If the derivative of f is given by f'(x)=ex-3x2, at which of the following values of x does f have a relative maximum value?

To find the relative maximum value of a function, we need to look for the critical points where the derivative is equal to zero or undefined.

Given that the derivative of f(x) is f'(x) = ex - 3x^2, we need to find the values of x for which f'(x) = 0.

Setting f'(x) = 0, we have:

ex - 3x^2 = 0

Now we solve this equation for x. Factoring out x:

x(e - 3x) = 0

From here, we have two possibilities:

1. x = 0
2. e - 3x = 0

For the second case, solving for x:

e = 3x
x = e/3

So, the possible values of x where f(x) may have a relative maximum are x = 0 and x = e/3.

To determine the values of x where the function f has a relative maximum value, we need to find the critical points of the function.

Step 1: Find the derivative:
Given f'(x) = ex - 3x^2.

Step 2: Set the derivative equal to zero and solve for x:
ex - 3x^2 = 0.

Step 3: Factor out the common term:
x(ex - 3x) = 0.

Step 4: Set each factor equal to zero and solve for x:
x = 0 or ex - 3x = 0.

To solve ex - 3x = 0, we can't find an exact algebraic solution, so we can use a numerical method or approximation technique like Newton's method or graphical analysis.

For now, let's focus on x = 0.

Step 5: Use the second derivative test:
To determine if x = 0 is a relative maximum, we need to evaluate the second derivative at x = 0.

The second derivative of f, denoted f''(x), is the derivative of f'(x).

f'(x) = ex - 3x^2.

Taking the derivative of f'(x) gives:
f''(x) = e^x - 6x.

Step 6: Evaluate the second derivative at x = 0:
f''(0) = e^0 - 6(0) = 1 - 0 = 1.

The second derivative at x = 0 is positive (1), which means the function has a relative minimum at x = 0, not a relative maximum.

Therefore, there is no relative maximum value of f at any of the given values of x.

set the derivative equal to zero and solve for x

ex - 3x^2 = 0
x(e - 3x) = 0
x = 0 or x = e/3

use the 2nd derivative test to see which produces the maximum

f ''(x) = e - 6x
f "(0) = e
f "(e/3) = e - 2e which is negative

so the value of x = e/3 produces a relative maximum value of the function.