Question

question 26
find all inflection points of the function: $g(x) = \frac{1}{4}x^4 - 6x^2 + 7$.
\bigcirc inflection points at $x = 0, 12$
\bigcirc inflection points at $x = -\sqrt{12}, 0, \sqrt{12}$
\bigcirc the graph of this function has no inflection points.
\bigcirc no correct answer choice is given.
\bigcirc inflection points at $x = -2, 2$

Explanation:

Step1: Find the second derivative

First, find the first derivative of \( g(x)=\frac{1}{4}x^{4}-6x^{2}+7 \). Using the power rule \( (x^n)' = nx^{n - 1} \), we have:
\( g'(x)=\frac{1}{4}\times4x^{3}-6\times2x=x^{3}-12x \)
Then, find the second derivative by differentiating \( g'(x) \):
\( g''(x)=3x^{2}-12 \)

Step2: Find critical points of the second derivative

Set \( g''(x) = 0 \) to find potential inflection points:
\( 3x^{2}-12 = 0 \)
Divide both sides by 3: \( x^{2}-4 = 0 \)
Factor: \( (x - 2)(x + 2)=0 \)
So, \( x = 2 \) or \( x=-2 \)

Step3: Test concavity around critical points

• For \( x < - 2 \), let's choose \( x=-3 \). Then \( g''(-3)=3\times(-3)^{2}-12=3\times9 - 12 = 27-12 = 15>0 \) (concave up)
• For \( -2 < x < 2 \), let's choose \( x = 0 \). Then \( g''(0)=3\times0^{2}-12=-12<0 \) (concave down)
• For \( x>2 \), let's choose \( x = 3 \). Then \( g''(3)=3\times3^{2}-12=27 - 12 = 15>0 \) (concave up)

Since the concavity changes at \( x=-2 \) and \( x = 2 \), these are inflection points.

Answer:

Inflection points at \( x=-2,2 \) (the option with this text)