The function f(x)=-2x^3+30x^2-96x+8 has one local minimum & one local maximum. This function has a local minimum at x equals ______ with value __________ and a local maximum at x equals _______ with value __________ .
f' = -6x^2 + 60x - 96
= -6(x^2-10x+16)
= -6(x-2)(x-8)
So, local min/max occurs where f' = 0
f" = -6(2x-10)
= -12(x-5)
f is max where f" < 0
f is min where f" > 0
Let 'er rip!
To find the local minimum and maximum of a function, we need to analyze the first and second derivatives of the function:
1. First Derivative:
The local minimum and maximum occur at the points where the first derivative of the function is equal to zero or does not exist.
Let's begin by finding the first derivative of the function f(x):
f'(x) = -6x^2 + 60x - 96
2. Critical Points (where f'(x) = 0):
To find the local minimum and maximum, we need to solve the equation f'(x) = 0 for x.
-6x^2 + 60x - 96 = 0
Now we can solve this quadratic equation to find the critical point(s).
Using factoring, we can divide the equation by -6 to simplify it:
x^2 - 10x + 16 = 0
The equation factors as:
(x - 2)(x - 8) = 0
Setting each factor to zero:
x - 2 = 0, so x = 2
x - 8 = 0, so x = 8
Hence, the critical points are x = 2 and x = 8.
3. Second Derivative:
To determine if the critical points are local minimum or maximum, we need to examine the second derivative of the function.
Let's find the second derivative:
f''(x) = -12x + 60
4. Analyzing the Second Derivative:
At x = 2:
f''(2) = -12(2) + 60 = 36
Since f''(2) > 0, the second derivative is positive, indicating a local minimum at x = 2.
At x = 8:
f''(8) = -12(8) + 60 = -36
Since f''(8) < 0, the second derivative is negative, indicating a local maximum at x = 8.
5. Local Minimum and Maximum:
From the analysis above, we found:
- Local Minimum: x = 2 with a value of f(2) = -2(2)^3 + 30(2)^2 - 96(2) + 8
- Local Maximum: x = 8 with a value of f(8) = -2(8)^3 + 30(8)^2 - 96(8) + 8
Simplifying these expressions will give you the final values for the local minimum and maximum of the function.