Math / Calculus

Question[-/0.5 Points] DETAILS MY NOTES TANAPMATH7 10.2.058.
Find the inflection point(s), if any, of the function. (If an answer does not exist, enter DNE.) $g(x)=4 x^{4}-8 x^{3}+6$ smaller $x$ -value $(x, y)=$ $\square$ ) larger $x$ -value $(x, y)=$ $\square$ ) Need Help? Read it

Studdy Solution

STEP 1

What is this asking? Find the points on the curve $g(x) = 4x^4 - 8x^3 + 6$ where the concavity changes, these are called inflection points! Watch out! Just because the second derivative is zero doesn't automatically mean it's an inflection point; the concavity actually needs to change around that point.

STEP 2

1. Find the second derivative.
2. Find possible inflection points.
3. Check for concavity changes.

STEP 3

$g'(x) = 16x^3 - 24x^2$ Remember, the power rule says we bring the exponent down and multiply it by the coefficient, then reduce the exponent by 1.

STEP 4

$g''(x) = 48x^2 - 48x$ We used the power rule again!

STEP 5

$48x^2 - 48x = 0$

STEP 6

$48x(x - 1) = 0$ This gives us two possible inflection points: $x = 0$ and $x = 1$ .
Exciting!

STEP 7

* For $x = -1$ : $g''(-1) = 48(-1)^2 - 48(-1) = 48 + 48 = 96$ .
Since $96 > 0$ , the function is concave up here. * For $x = 0.5$ : $g''(0.5) = 48(0.5)^2 - 48(0.5) = 12 - 24 = -12$ .
Since $-12 < 0$ , the function is concave down here. * For $x = 2$ : $g''(2) = 48(2)^2 - 48(2) = 192 - 96 = 96$ .
Since $96 > 0$ , the function is concave up here.

STEP 8

* For $x = 0$ : $g(0) = 4(0)^4 - 8(0)^3 + 6 = 6$ .
So, the first inflection point is $(0, 6)$ . * For $x = 1$ : $g(1) = 4(1)^4 - 8(1)^3 + 6 = 4 - 8 + 6 = 2$ .
So, the second inflection point is $(1, 2)$ .

STEP 9

Smaller $x$ -value $(x, y) = (0, 6)$ Larger $x$ -value $(x, y) = (1, 2)$

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Question[-/0.5 Points] DETAILS MY NOTES TANAPMATH7 10.2.058.
Find the inflection point(s), if any, of the function. (If an answer does not exist, enter DNE.) $g(x)=4 x^{4}-8 x^{3}+6$ smaller $x$ -value $(x, y)=$ $\square$ ) larger $x$ -value $(x, y)=$ $\square$ ) Need Help? Read it

Studdy Solution

STEP 1

STEP 2

1. Find the second derivative.
2. Find possible inflection points.
3. Check for concavity changes.

STEP 3

$g'(x) = 16x^3 - 24x^2$ Remember, the power rule says we bring the exponent down and multiply it by the coefficient, then reduce the exponent by 1.

STEP 4

$g''(x) = 48x^2 - 48x$ We used the power rule again!

STEP 5

$48x^2 - 48x = 0$

STEP 6

$48x(x - 1) = 0$ This gives us two possible inflection points: $x = 0$ and $x = 1$ .
Exciting!

STEP 7

STEP 8

* For $x = 0$ : $g(0) = 4(0)^4 - 8(0)^3 + 6 = 6$ .
So, the first inflection point is $(0, 6)$ . * For $x = 1$ : $g(1) = 4(1)^4 - 8(1)^3 + 6 = 4 - 8 + 6 = 2$ .
So, the second inflection point is $(1, 2)$ .

STEP 9

Smaller $x$ -value $(x, y) = (0, 6)$ Larger $x$ -value $(x, y) = (1, 2)$

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Studdy Solution

STEP 1

STEP 2

1. Find the second derivative.2. Find possible inflection points.3. Check for concavity changes.

STEP 3

g′(x)=16x3−24x2g'(x) = 16x^3 - 24x^2g′(x)=16x3−24x2 Remember, the power rule says we bring the exponent down and multiply it by the coefficient, then reduce the exponent by 1.

STEP 4

g′′(x)=48x2−48xg''(x) = 48x^2 - 48xg′′(x)=48x2−48x We used the power rule again!

STEP 5

48x2−48x=048x^2 - 48x = 048x2−48x=0

STEP 6

48x(x−1)=048x(x - 1) = 048x(x−1)=0 This gives us two possible inflection points: x=0x = 0x=0 and x=1x = 1x=1.Exciting!

STEP 7

STEP 8

STEP 9

Smaller xxx-value (x,y)=(0,6)(x, y) = (0, 6)(x,y)=(0,6) Larger xxx-value (x,y)=(1,2)(x, y) = (1, 2)(x,y)=(1,2)

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1. Find the second derivative.
2. Find possible inflection points.
3. Check for concavity changes.

$g'(x) = 16x^3 - 24x^2$ Remember, the power rule says we bring the exponent down and multiply it by the coefficient, then reduce the exponent by 1.

$g''(x) = 48x^2 - 48x$ We used the power rule again!

$48x^2 - 48x = 0$

$48x(x - 1) = 0$ This gives us two possible inflection points: $x = 0$ and $x = 1$ .
Exciting!

Smaller $x$ -value $(x, y) = (0, 6)$ Larger $x$ -value $(x, y) = (1, 2)$