Answer :
We start with the function
[tex]$$
f(x) = -12x^5 - 45x^4 + 80x^3 - 8.
$$[/tex]
A critical number is a number [tex]$x$[/tex] where the derivative [tex]$f'(x)$[/tex] is either zero or undefined. For this function, the derivative exists for all real numbers, so we only need to solve
[tex]$$
f'(x) = 0.
$$[/tex]
Step 1. Find the first derivative:
Differentiate term by term:
- The derivative of [tex]$-12x^5$[/tex] is [tex]$-60x^4$[/tex].
- The derivative of [tex]$-45x^4$[/tex] is [tex]$-180x^3$[/tex].
- The derivative of [tex]$80x^3$[/tex] is [tex]$240x^2$[/tex].
- The derivative of the constant [tex]$-8$[/tex] is [tex]$0$[/tex].
Thus, we have
[tex]$$
f'(x) = -60x^4 - 180x^3 + 240x^2.
$$[/tex]
Step 2. Factor the derivative:
Notice that there is a common factor in all terms. In fact, we can factor out [tex]$-60x^2$[/tex]:
[tex]$$
f'(x) = -60x^2(x^2 + 3x - 4).
$$[/tex]
Now, factor the quadratic [tex]$x^2 + 3x - 4$[/tex]. We look for two numbers whose product is [tex]$-4$[/tex] and whose sum is [tex]$3$[/tex]. These numbers are [tex]$4$[/tex] and [tex]$-1$[/tex]. Therefore,
[tex]$$
x^2 + 3x - 4 = (x + 4)(x - 1).
$$[/tex]
The derivative can now be written as
[tex]$$
f'(x) = -60x^2 (x + 4)(x - 1).
$$[/tex]
Step 3. Solve [tex]$f'(x) = 0$[/tex]:
Set each factor equal to zero:
1. [tex]$-60$[/tex] is never zero.
2. [tex]$x^2 = 0$[/tex] gives [tex]$x = 0$[/tex].
3. [tex]$x + 4 = 0$[/tex] gives [tex]$x = -4$[/tex].
4. [tex]$x - 1 = 0$[/tex] gives [tex]$x = 1$[/tex].
So, the critical numbers are
[tex]$$
x = -4,\quad x = 0,\quad x = 1.
$$[/tex]
Step 4. Classify each critical number using the second derivative:
Compute the second derivative [tex]$f''(x)$[/tex] by differentiating [tex]$f'(x)$[/tex]:
Starting with
[tex]$$
f'(x) = -60x^4 - 180x^3 + 240x^2,
$$[/tex]
differentiate term by term:
- The derivative of [tex]$-60x^4$[/tex] is [tex]$-240x^3$[/tex].
- The derivative of [tex]$-180x^3$[/tex] is [tex]$-540x^2$[/tex].
- The derivative of [tex]$240x^2$[/tex] is [tex]$480x$[/tex].
Thus,
[tex]$$
f''(x) = -240x^3 - 540x^2 + 480x.
$$[/tex]
Now evaluate [tex]$f''(x)$[/tex] at each critical number.
1. At [tex]$x = -4$[/tex]:
[tex]$$
f''(-4) = -240(-4)^3 - 540(-4)^2 + 480(-4).
$$[/tex]
Computing the sign (without doing the full arithmetic) shows that [tex]$f''(-4)$[/tex] is positive. Therefore, [tex]$x = -4$[/tex] is a local minimum.
2. At [tex]$x = 0$[/tex]:
[tex]$$
f''(0) = -240(0)^3 - 540(0)^2 + 480(0) = 0.
$$[/tex]
Since [tex]$f''(0) = 0$[/tex], the second derivative test is inconclusive. This suggests that [tex]$x = 0$[/tex] is an inflection point or a point where the test does not determine a maximum or minimum.
3. At [tex]$x = 1$[/tex]:
[tex]$$
f''(1) = -240(1)^3 - 540(1)^2 + 480(1) = -240 - 540 + 480 = -300.
$$[/tex]
Since [tex]$f''(1)$[/tex] is negative, the function is concave down at [tex]$x = 1$[/tex], meaning that [tex]$x = 1$[/tex] is a local maximum.
Step 5. Summary of results:
- The critical number [tex]$x = -4$[/tex] is a local minimum.
- The critical number [tex]$x = 0$[/tex] is an inflection point (the second derivative test is inconclusive).
- The critical number [tex]$x = 1$[/tex] is a local maximum.
Thus, the final answer is:
[tex]$$
\text{Critical numbers: } x = -4,\quad x = 0,\quad x = 1.
$$[/tex]
With the classifications:
- [tex]$x = -4$[/tex] is a local minimum.
- [tex]$x = 0$[/tex] is an inflection point (test inconclusive).
- [tex]$x = 1$[/tex] is a local maximum.
[tex]$$
f(x) = -12x^5 - 45x^4 + 80x^3 - 8.
$$[/tex]
A critical number is a number [tex]$x$[/tex] where the derivative [tex]$f'(x)$[/tex] is either zero or undefined. For this function, the derivative exists for all real numbers, so we only need to solve
[tex]$$
f'(x) = 0.
$$[/tex]
Step 1. Find the first derivative:
Differentiate term by term:
- The derivative of [tex]$-12x^5$[/tex] is [tex]$-60x^4$[/tex].
- The derivative of [tex]$-45x^4$[/tex] is [tex]$-180x^3$[/tex].
- The derivative of [tex]$80x^3$[/tex] is [tex]$240x^2$[/tex].
- The derivative of the constant [tex]$-8$[/tex] is [tex]$0$[/tex].
Thus, we have
[tex]$$
f'(x) = -60x^4 - 180x^3 + 240x^2.
$$[/tex]
Step 2. Factor the derivative:
Notice that there is a common factor in all terms. In fact, we can factor out [tex]$-60x^2$[/tex]:
[tex]$$
f'(x) = -60x^2(x^2 + 3x - 4).
$$[/tex]
Now, factor the quadratic [tex]$x^2 + 3x - 4$[/tex]. We look for two numbers whose product is [tex]$-4$[/tex] and whose sum is [tex]$3$[/tex]. These numbers are [tex]$4$[/tex] and [tex]$-1$[/tex]. Therefore,
[tex]$$
x^2 + 3x - 4 = (x + 4)(x - 1).
$$[/tex]
The derivative can now be written as
[tex]$$
f'(x) = -60x^2 (x + 4)(x - 1).
$$[/tex]
Step 3. Solve [tex]$f'(x) = 0$[/tex]:
Set each factor equal to zero:
1. [tex]$-60$[/tex] is never zero.
2. [tex]$x^2 = 0$[/tex] gives [tex]$x = 0$[/tex].
3. [tex]$x + 4 = 0$[/tex] gives [tex]$x = -4$[/tex].
4. [tex]$x - 1 = 0$[/tex] gives [tex]$x = 1$[/tex].
So, the critical numbers are
[tex]$$
x = -4,\quad x = 0,\quad x = 1.
$$[/tex]
Step 4. Classify each critical number using the second derivative:
Compute the second derivative [tex]$f''(x)$[/tex] by differentiating [tex]$f'(x)$[/tex]:
Starting with
[tex]$$
f'(x) = -60x^4 - 180x^3 + 240x^2,
$$[/tex]
differentiate term by term:
- The derivative of [tex]$-60x^4$[/tex] is [tex]$-240x^3$[/tex].
- The derivative of [tex]$-180x^3$[/tex] is [tex]$-540x^2$[/tex].
- The derivative of [tex]$240x^2$[/tex] is [tex]$480x$[/tex].
Thus,
[tex]$$
f''(x) = -240x^3 - 540x^2 + 480x.
$$[/tex]
Now evaluate [tex]$f''(x)$[/tex] at each critical number.
1. At [tex]$x = -4$[/tex]:
[tex]$$
f''(-4) = -240(-4)^3 - 540(-4)^2 + 480(-4).
$$[/tex]
Computing the sign (without doing the full arithmetic) shows that [tex]$f''(-4)$[/tex] is positive. Therefore, [tex]$x = -4$[/tex] is a local minimum.
2. At [tex]$x = 0$[/tex]:
[tex]$$
f''(0) = -240(0)^3 - 540(0)^2 + 480(0) = 0.
$$[/tex]
Since [tex]$f''(0) = 0$[/tex], the second derivative test is inconclusive. This suggests that [tex]$x = 0$[/tex] is an inflection point or a point where the test does not determine a maximum or minimum.
3. At [tex]$x = 1$[/tex]:
[tex]$$
f''(1) = -240(1)^3 - 540(1)^2 + 480(1) = -240 - 540 + 480 = -300.
$$[/tex]
Since [tex]$f''(1)$[/tex] is negative, the function is concave down at [tex]$x = 1$[/tex], meaning that [tex]$x = 1$[/tex] is a local maximum.
Step 5. Summary of results:
- The critical number [tex]$x = -4$[/tex] is a local minimum.
- The critical number [tex]$x = 0$[/tex] is an inflection point (the second derivative test is inconclusive).
- The critical number [tex]$x = 1$[/tex] is a local maximum.
Thus, the final answer is:
[tex]$$
\text{Critical numbers: } x = -4,\quad x = 0,\quad x = 1.
$$[/tex]
With the classifications:
- [tex]$x = -4$[/tex] is a local minimum.
- [tex]$x = 0$[/tex] is an inflection point (test inconclusive).
- [tex]$x = 1$[/tex] is a local maximum.
