Math / Calculus

QuestionConsider the function. (If an answer does not exist, enter DNE.) $f(x)=(x-4)^{2}(x-8)^{2}$ (a) Determine intervals where $f$ is increasing or decreasing. (Enter your answers using interval notation.) increasing $\quad(4,6) \cup(8, \infty)$ decreasing $\quad(-\infty, 4) \cup(6,8)$ (b) Determine the local minima and maxima of $f$ . (Enter your answers as comma-separated lists.) locations of local minima $\quad x=$ $\square$ $\square$ locations of local maxima $\quad x=$ $\square$ (c) Determine intervals where $f$ is concave up or concave down. (Enter your answers using interval notation.) concave up $\quad(-\infty, 4.84) \cup(7.16, \infty)$ concave down (4.84,7.16) (d) Determine the locations of inflection points of $f$ . Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. (Enter your answers as a comma-separated list.) $x=4.84,7.16$

Studdy Solution

STEP 1

1. The function $f(x) = (x-4)^2(x-8)^2$ is a polynomial function.
2. To determine intervals of increase and decrease, we need the first derivative.
3. To find local minima and maxima, we analyze critical points from the first derivative.
4. To determine concavity, we need the second derivative.
5. Inflection points occur where the second derivative changes sign.

STEP 2

1. Find the first derivative $f'(x)$ .
2. Determine intervals of increase and decrease using $f'(x)$ .
3. Find critical points and determine local minima and maxima.
4. Find the second derivative $f''(x)$ .
5. Determine intervals of concavity using $f''(x)$ .
6. Determine inflection points.

STEP 3

Find the first derivative $f'(x)$ using the product rule and chain rule:
$f(x) = (x-4)^2(x-8)^2$
Using the product rule:
$f'(x) = \frac{d}{dx}[(x-4)^2] \cdot (x-8)^2 + (x-4)^2 \cdot \frac{d}{dx}[(x-8)^2]$
Calculate derivatives:
$\frac{d}{dx}[(x-4)^2] = 2(x-4)$ $\frac{d}{dx}[(x-8)^2] = 2(x-8)$
Substitute back:
$f'(x) = 2(x-4)(x-8)^2 + (x-4)^2 \cdot 2(x-8)$
Simplify:
$f'(x) = 2(x-4)(x-8)^2 + 2(x-4)^2(x-8)$
Factor out common terms:
$f'(x) = 2(x-4)(x-8)[(x-8) + (x-4)]$
$f'(x) = 2(x-4)(x-8)(2x-12)$
$f'(x) = 4(x-4)(x-8)(x-6)$

STEP 4

Determine intervals of increase and decrease by analyzing $f'(x) = 4(x-4)(x-8)(x-6)$ :
Find critical points by setting $f'(x) = 0$ :
$4(x-4)(x-8)(x-6) = 0$
Critical points are $x = 4, 6, 8$ .
Test intervals around these points:
- $(-\infty, 4)$ : Choose $x = 0$ , $f'(0) < 0$ , so $f$ is decreasing. - $(4, 6)$ : Choose $x = 5$ , $f'(5) > 0$ , so $f$ is increasing. - $(6, 8)$ : Choose $x = 7$ , $f'(7) < 0$ , so $f$ is decreasing. - $(8, \infty)$ : Choose $x = 9$ , $f'(9) > 0$ , so $f$ is increasing.
Intervals: - Increasing: $(4, 6) \cup (8, \infty)$ - Decreasing: $(-\infty, 4) \cup (6, 8)$

STEP 5

Determine local minima and maxima:
- Local minima occur where $f'(x)$ changes from negative to positive. - Local maxima occur where $f'(x)$ changes from positive to negative.
From the intervals: - Local minima at $x = 4$ and $x = 8$ . - Local maxima at $x = 6$ .

STEP 6

Find the second derivative $f''(x)$ :
Differentiate $f'(x) = 4(x-4)(x-8)(x-6)$ using the product rule and simplify:
$f''(x) = \frac{d}{dx}[4(x-4)(x-8)(x-6)]$
Apply product rule and simplify:
$f''(x) = 4[(x-8)(x-6) + (x-4)(x-6) + (x-4)(x-8)]$
$f''(x) = 4[(x^2 - 14x + 48) + (x^2 - 10x + 24) + (x^2 - 12x + 32)]$
$f''(x) = 4(3x^2 - 36x + 104)$

STEP 7

Determine intervals of concavity by analyzing $f''(x) = 4(3x^2 - 36x + 104)$ :
Find where $f''(x) = 0$ :
$3x^2 - 36x + 104 = 0$
Use the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{36 \pm \sqrt{(-36)^2 - 4 \cdot 3 \cdot 104}}{2 \cdot 3}$
$x = \frac{36 \pm \sqrt{1296 - 1248}}{6}$
$x = \frac{36 \pm \sqrt{48}}{6}$
$x = \frac{36 \pm 4\sqrt{3}}{6}$
$x = 6 \pm \frac{2\sqrt{3}}{3}$
Approximate solutions:
$x \approx 4.84, 7.16$
Test intervals around these points:
- Concave up: $(-\infty, 4.84) \cup (7.16, \infty)$ - Concave down: $(4.84, 7.16)$

STEP 8

Determine inflection points where $f''(x)$ changes sign:
Inflection points at $x \approx 4.84, 7.16$ .
The solutions are: (a) Increasing: $(4, 6) \cup (8, \infty)$ , Decreasing: $(-\infty, 4) \cup (6, 8)$ (b) Local minima: $x = 4, 8$ , Local maxima: $x = 6$ (c) Concave up: $(-\infty, 4.84) \cup (7.16, \infty)$ , Concave down: $(4.84, 7.16)$ (d) Inflection points: $x = 4.84, 7.16$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms