Find the absolute minimum and absolute maximum values of f on the interval below.

f(x) = x4 - 8x2 + 6
[-3, 4]

f = x^4 - 8x^2 + 6

f' = 4x^3 - 16x
f" = 12x^2 - 16

f'=0 at x = 0,±2
at x = -2, f" > 0 so f has a local min
at x = 0, f" = < 0 so f has a local max
at x = 2, f" > 0 so f has a local min

So, now just evaluate f(x) at 0,±2 and the endpoints of the interval, to see what the absolute extrema are for the interval.

good

To find the absolute minimum and absolute maximum values of the function f(x) = x^4 - 8x^2 + 6 on the interval [-3, 4], we need to consider the critical points and the endpoints of the interval.

1. Critical points:
To find the critical points, we need to find the values of x where the derivative of f(x) is equal to zero or is undefined.
First, let's find the derivative of f(x):
f'(x) = 4x^3 - 16x

Now, set f'(x) = 0 and solve for x:
4x^3 - 16x = 0
4x(x^2 - 4) = 0
x(x + 2)(x - 2) = 0

So, the critical points are x = -2, 0, and 2.

2. Endpoints of the interval:
The interval [-3, 4] has endpoints x = -3 and x = 4, so we need to evaluate f(x) at these points as well.

Now, plug in the critical points and the endpoints into the original function f(x) to find the corresponding values:

f(-3) = (-3)^4 - 8(-3)^2 + 6 = 81 - 72 + 6 = 15
f(-2) = (-2)^4 - 8(-2)^2 + 6 = 16 - 32 + 6 = -10
f(0) = 0^4 - 8(0)^2 + 6 = 6
f(2) = 2^4 - 8(2)^2 + 6 = 16 - 32 + 6 = -10
f(4) = 4^4 - 8(4)^2 + 6 = 256 - 128 + 6 = 134

So, the values of f(x) at the critical points and the endpoints are:
f(-3) = 15
f(-2) = -10
f(0) = 6
f(2) = -10
f(4) = 134

The absolute maximum value of f(x) on the interval [-3, 4] is 134, and it occurs at x = 4.
The absolute minimum value of f(x) on the interval [-3, 4] is -10, and it occurs at x = -2 and x = 2.

To find the absolute minimum and absolute maximum values of a function on an interval, we can follow these steps:

1. Find the critical points of the function within the interval.
2. Evaluate the function at these critical points and at the endpoints of the interval.
3. Compare the values obtained in step 2 to determine the absolute minimum and absolute maximum.

Let's apply these steps to find the absolute minimum and absolute maximum values of the function f(x) = x^4 - 8x^2 + 6 on the interval [-3, 4].

Step 1: Find the critical points.
To find the critical points, we need to find where the derivative of the function f(x) is equal to zero or undefined.

First, let's find the derivative of f(x):
f'(x) = 4x^3 - 16x

Now, let's set f'(x) = 0 and solve for x:
4x^3 - 16x = 0
4x(x^2 - 4) = 0
x(x + 2)(x - 2) = 0

This equation is satisfied when x = 0, x = -2, and x = 2. So these are the critical points within the given interval.

Step 2: Evaluate the function at the critical points and endpoints.
Now, let's evaluate the function f(x) at the critical points and endpoints:

For x = -3:
f(-3) = (-3)^4 - 8(-3)^2 + 6 = 81 - 72 + 6 = 15

For x = 4:
f(4) = (4)^4 - 8(4)^2 + 6 = 256 - 128 + 6 = 134

For x = -2:
f(-2) = (-2)^4 - 8(-2)^2 + 6 = 16 - 32 + 6 = -10

For x = 2:
f(2) = (2)^4 - 8(2)^2 + 6 = 16 - 32 + 6 = -10

For x = 0:
f(0) = (0)^4 - 8(0)^2 + 6 = 0 - 0 + 6 = 6

Step 3: Compare the values obtained.
Comparing the values obtained, we find:
The absolute minimum value of f(x) on the interval [-3, 4] is -10, which occurs at x = -2 and x = 2.
The absolute maximum value of f(x) on the interval [-3, 4] is 134, which occurs at x = 4.

Therefore, the absolute minimum value is -10 and the absolute maximum value is 134 for the function f(x) = x^4 - 8x^2 + 6 on the interval [-3, 4].