Consider the function f(x)=(x^3)(e^9x),
-2 is less than or equal to x is less than or equal to 4
*The absolute maximum value is _____ and this occurs at x equals 4.
*The absolute minimum value is _____ and this occurs at x equals ______
I got the 4 but I must be doing something wrong with the other numbers because it is not working
let's take the derivative
f'(x) = 3x^2 (e^(9x) ) + x^3 (9) e^(9x)
= 3x^2 e^(9x) [ 1 + 3x]
= 0 for a max/min
x = 0 , x = -1/3
f(0) = 0
f(-1/3) = (-1/27) e^-3 = -1/(27e^3) = appr -.00184
f(-2) = -8 e^-18 = -8/e^18 = appr -.0000000121
so the absolute minimum is -8/e^18 , when x = -2
wrong is x=-4
To find the absolute maximum and minimum values of the function f(x)=(x^3)(e^9x) on the interval [-2, 4], you can use the following steps:
Step 1: Find all critical points by taking the derivative of f(x) and setting it equal to zero:
f'(x) = 3x^2 * (e^9x) + x^3 * 9(e^9x) = 0
Step 2: Solve the equation obtained in step 1 for x to find the critical points.
3x^2 * (e^9x) + x^3 * 9(e^9x) = 0
e^9x * [(3x^2 + 9x^3)] = 0
Either e^9x = 0 (which is not possible) or 3x^2 + 9x^3 = 0
Factoring out the common term:
3x^2(1 + 3x) = 0
This equation has two solutions: x = 0 and x = -1/3
Step 3: Evaluate the function at the critical points and the endpoints of the interval.
f(-2) = (-2)^3 * (e^9(-2)) ≈ -2.12
f(0) = (0)^3 * (e^9(0)) = 0
f(-1/3) = (-1/3)^3 * (e^9(-1/3)) ≈ 0.03
f(4) = (4)^3 * (e^9(4)) ≈ 150,747.81
Step 4: Compare the function values obtained in step 3 to determine the absolute maximum and minimum values.
From the values, f(-2), f(0), f(-1/3), f(4), we can say:
The absolute maximum value is approximately 150,747.81, and it occurs at x = 4.
The absolute minimum value is approximately -2.12, and it occurs at x = -2.
Make sure to double-check your calculations to ensure accuracy.