Answer :
To find the critical numbers of the given functions and classify each critical point as a relative maximum, a relative minimum, or neither, let's break down the process step-by-step.
For problem 23, we have the function:
[tex]\[ f(x) = 3x^4 - 8x^3 + 6x^2 + 2 \][/tex]
1. Find the first derivative:
The first derivative, [tex]\( f'(x) \)[/tex], is used to find the critical points. Calculate it:
[tex]\[
f'(x) = 12x^3 - 24x^2 + 12x
\][/tex]
2. Find the critical points:
Set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[
12x^3 - 24x^2 + 12x = 0
\][/tex]
Factor the expression:
[tex]\[
12x(x^2 - 2x + 1) = 0
\][/tex]
Further factor:
[tex]\[
12x(x-1)^2 = 0
\][/tex]
So, the critical points are [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex].
3. Classify the critical points using the second derivative test:
Find the second derivative:
[tex]\[
f''(x) = 36x^2 - 48x + 12
\][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[
f''(0) = 36(0)^2 - 48(0) + 12 = 12 \quad (\text{positive, so relative minimum})
\][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[
f''(1) = 36(1)^2 - 48(1) + 12 = 0 \quad (\text{neither maximum nor minimum})
\][/tex]
Thus, for problem 23, the critical points are [tex]\( x = 0 \)[/tex] (relative minimum) and [tex]\( x = 1 \)[/tex] (neither).
For problem 24, we have the function:
[tex]\[ f(x) = 324x - 72x^2 + 4x^3 \][/tex]
1. Find the first derivative:
Calculate the first derivative:
[tex]\[
f'(x) = 324 - 144x + 12x^2
\][/tex]
2. Find the critical points:
Set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[
12x^2 - 144x + 324 = 0
\][/tex]
Simplify the equation:
[tex]\[
x^2 - 12x + 27 = 0
\][/tex]
Factor or use the quadratic formula to find the roots:
[tex]\[
(x - 3)(x - 9) = 0
\][/tex]
So, the critical points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 9 \)[/tex].
3. Classify the critical points using the second derivative test:
Find the second derivative:
[tex]\[
f''(x) = 24x - 144
\][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[
f''(3) = 24(3) - 144 = -72 \quad (\text{negative, so relative maximum})
\][/tex]
- For [tex]\( x = 9 \)[/tex]:
[tex]\[
f''(9) = 24(9) - 144 = 72 \quad (\text{positive, so relative minimum})
\][/tex]
Thus, for problem 24, the critical points are [tex]\( x = 3 \)[/tex] (relative maximum) and [tex]\( x = 9 \)[/tex] (relative minimum).
For problem 23, we have the function:
[tex]\[ f(x) = 3x^4 - 8x^3 + 6x^2 + 2 \][/tex]
1. Find the first derivative:
The first derivative, [tex]\( f'(x) \)[/tex], is used to find the critical points. Calculate it:
[tex]\[
f'(x) = 12x^3 - 24x^2 + 12x
\][/tex]
2. Find the critical points:
Set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[
12x^3 - 24x^2 + 12x = 0
\][/tex]
Factor the expression:
[tex]\[
12x(x^2 - 2x + 1) = 0
\][/tex]
Further factor:
[tex]\[
12x(x-1)^2 = 0
\][/tex]
So, the critical points are [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex].
3. Classify the critical points using the second derivative test:
Find the second derivative:
[tex]\[
f''(x) = 36x^2 - 48x + 12
\][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[
f''(0) = 36(0)^2 - 48(0) + 12 = 12 \quad (\text{positive, so relative minimum})
\][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[
f''(1) = 36(1)^2 - 48(1) + 12 = 0 \quad (\text{neither maximum nor minimum})
\][/tex]
Thus, for problem 23, the critical points are [tex]\( x = 0 \)[/tex] (relative minimum) and [tex]\( x = 1 \)[/tex] (neither).
For problem 24, we have the function:
[tex]\[ f(x) = 324x - 72x^2 + 4x^3 \][/tex]
1. Find the first derivative:
Calculate the first derivative:
[tex]\[
f'(x) = 324 - 144x + 12x^2
\][/tex]
2. Find the critical points:
Set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[
12x^2 - 144x + 324 = 0
\][/tex]
Simplify the equation:
[tex]\[
x^2 - 12x + 27 = 0
\][/tex]
Factor or use the quadratic formula to find the roots:
[tex]\[
(x - 3)(x - 9) = 0
\][/tex]
So, the critical points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 9 \)[/tex].
3. Classify the critical points using the second derivative test:
Find the second derivative:
[tex]\[
f''(x) = 24x - 144
\][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[
f''(3) = 24(3) - 144 = -72 \quad (\text{negative, so relative maximum})
\][/tex]
- For [tex]\( x = 9 \)[/tex]:
[tex]\[
f''(9) = 24(9) - 144 = 72 \quad (\text{positive, so relative minimum})
\][/tex]
Thus, for problem 24, the critical points are [tex]\( x = 3 \)[/tex] (relative maximum) and [tex]\( x = 9 \)[/tex] (relative minimum).
