how do you find the absolute maximum of 2x^3+3x^2-12x-7 on interval (-3,0)
To find the absolute maximum of a function on a given interval, you need to follow these steps:
1. Find the critical points: Critical points occur where the derivative of the function equals zero or is undefined. Start by taking the derivative of the given function.
For the function f(x) = 2x^3 + 3x^2 - 12x - 7, the derivative is f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2).
Set f'(x) = 0 and solve for x:
6(x^2 + x - 2) = 0
x^2 + x - 2 = 0
(x + 2)(x - 1) = 0
So, the critical points are x = -2 and x = 1.
2. Determine endpoints: Check the function value at both endpoints of the given interval, -3 and 0.
f(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) - 7 = -17
f(0) = 2(0)^3 + 3(0)^2 - 12(0) - 7 = -7
3. Evaluate the function value at the critical points: Calculate f(x) at the critical points we found in step 1.
f(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) - 7 = 7
f(1) = 2(1)^3 + 3(1)^2 - 12(1) - 7 = -14
4. Compare the function values: Compare all the function values obtained in steps 2 and 3. The highest function value will be the absolute maximum.
The function values are:
f(-3) = -17
f(-2) = 7
f(0) = -7
f(1) = -14
Thus, the absolute maximum is 7 and it occurs at x = -2.
Therefore, the absolute maximum of the function 2x^3 + 3x^2 - 12x - 7 on the interval (-3, 0) is 7, and it occurs at x = -2.