Math / Calculus

QuestionConsider the following. (If an answer does not exist, enter DNE.) $f(x)=x^{3}-3 x^{2}-7 x+5$
Find the interval(s) on which $f$ is concave up. (Enter your answer using interval notation.) $\square$ Find the interval(s) on which $f$ is concave down. (Enter your answer using interval notation.) $\square$ Find the inflection point of $f$ . $\square$ $(x, y)=(\square)$ Need Help? Read It

Studdy Solution

STEP 1

1. The function given is $f(x) = x^3 - 3x^2 - 7x + 5$ .
2. Concavity is determined by the second derivative of the function.
3. An inflection point occurs where the concavity changes, which is where the second derivative is zero and changes sign.

STEP 2

1. Find the first derivative of $f(x)$ .
2. Find the second derivative of $f(x)$ .
3. Determine where the second derivative is zero or undefined.
4. Test intervals around these points to determine concavity.
5. Identify intervals of concave up and concave down.
6. Find the inflection point(s).

STEP 3

Find the first derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 7x + 5)$
$f'(x) = 3x^2 - 6x - 7$

STEP 4

Find the second derivative of $f(x)$ .
$f''(x) = \frac{d}{dx}(3x^2 - 6x - 7)$
$f''(x) = 6x - 6$

STEP 5

Determine where the second derivative is zero or undefined.
$6x - 6 = 0$
Solve for $x$ :
$6x = 6$
$x = 1$

STEP 6

Test intervals around $x = 1$ to determine concavity.
Choose a test point in each interval: - For $x < 1$ , choose $x = 0$ . - For $x > 1$ , choose $x = 2$ .
Calculate $f''(x)$ at these points: - $f''(0) = 6(0) - 6 = -6$ (Concave down) - $f''(2) = 6(2) - 6 = 6$ (Concave up)

STEP 7

Identify intervals of concave up and concave down.
- Concave up: $(1, \infty)$ - Concave down: $(-\infty, 1)$

STEP 8

Find the inflection point(s), where the concavity changes.
Since $x = 1$ is where $f''(x)$ changes sign, it is an inflection point.
Calculate $f(1)$ :
$f(1) = (1)^3 - 3(1)^2 - 7(1) + 5$
$f(1) = 1 - 3 - 7 + 5 = -4$
The inflection point is $(1, -4)$ .
The intervals and inflection point are: - Concave up: $(1, \infty)$ - Concave down: $(-\infty, 1)$ - Inflection point: $(1, -4)$

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Studdy Solution

STEP 1

1. The function given is $f(x) = x^3 - 3x^2 - 7x + 5$ .
2. Concavity is determined by the second derivative of the function.
3. An inflection point occurs where the concavity changes, which is where the second derivative is zero and changes sign.

STEP 2

1. Find the first derivative of $f(x)$ .
2. Find the second derivative of $f(x)$ .
3. Determine where the second derivative is zero or undefined.
4. Test intervals around these points to determine concavity.
5. Identify intervals of concave up and concave down.
6. Find the inflection point(s).

STEP 3

Find the first derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 7x + 5)$
$f'(x) = 3x^2 - 6x - 7$

STEP 4

Find the second derivative of $f(x)$ .
$f''(x) = \frac{d}{dx}(3x^2 - 6x - 7)$
$f''(x) = 6x - 6$

STEP 5

Determine where the second derivative is zero or undefined.
$6x - 6 = 0$
Solve for $x$ :
$6x = 6$
$x = 1$

STEP 6

Test intervals around $x = 1$ to determine concavity.
Choose a test point in each interval: - For $x < 1$ , choose $x = 0$ . - For $x > 1$ , choose $x = 2$ .
Calculate $f''(x)$ at these points: - $f''(0) = 6(0) - 6 = -6$ (Concave down) - $f''(2) = 6(2) - 6 = 6$ (Concave up)

STEP 7

Identify intervals of concave up and concave down.
- Concave up: $(1, \infty)$ - Concave down: $(-\infty, 1)$

STEP 8

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Studdy Solution

STEP 1

1. The function given is f(x)=x3−3x2−7x+5 f(x) = x^3 - 3x^2 - 7x + 5 f(x)=x3−3x2−7x+5.2. Concavity is determined by the second derivative of the function.3. An inflection point occurs where the concavity changes, which is where the second derivative is zero and changes sign.

STEP 2

STEP 3

Find the first derivative of f(x) f(x) f(x).f′(x)=ddx(x3−3x2−7x+5) f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 7x + 5) f′(x)=dxd​(x3−3x2−7x+5)f′(x)=3x2−6x−7 f'(x) = 3x^2 - 6x - 7 f′(x)=3x2−6x−7

STEP 4

Find the second derivative of f(x) f(x) f(x).f′′(x)=ddx(3x2−6x−7) f''(x) = \frac{d}{dx}(3x^2 - 6x - 7) f′′(x)=dxd​(3x2−6x−7)f′′(x)=6x−6 f''(x) = 6x - 6 f′′(x)=6x−6

STEP 5

Determine where the second derivative is zero or undefined.6x−6=0 6x - 6 = 0 6x−6=0Solve for x x x:6x=6 6x = 6 6x=6x=1 x = 1 x=1

STEP 6

STEP 7

Identify intervals of concave up and concave down.- Concave up: (1,∞) (1, \infty) (1,∞) - Concave down: (−∞,1) (-\infty, 1) (−∞,1)

STEP 8

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1. The function given is $f(x) = x^3 - 3x^2 - 7x + 5$ .
2. Concavity is determined by the second derivative of the function.
3. An inflection point occurs where the concavity changes, which is where the second derivative is zero and changes sign.

Find the first derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 7x + 5)$
$f'(x) = 3x^2 - 6x - 7$

Find the second derivative of $f(x)$ .
$f''(x) = \frac{d}{dx}(3x^2 - 6x - 7)$
$f''(x) = 6x - 6$

Determine where the second derivative is zero or undefined.
$6x - 6 = 0$
Solve for $x$ :
$6x = 6$
$x = 1$

Identify intervals of concave up and concave down.
- Concave up: $(1, \infty)$ - Concave down: $(-\infty, 1)$