Given the function $g(x) = 3x^4 - 2x^2 + 49$, we need to find the first and second derivatives, $g'(x)$ and $g''(x)$. Then, we need to analyze the sign of $g'(x)$ and $g''(x)$, and finally, find the value of $x$ for which $g''(x) = 0$.

1. Problem Description

Given the function $g(x) = 3x^4 - 2x^2 + 49$, we need to find the first and second derivatives, $g'(x)$ and $g''(x)$. Then, we need to analyze the sign of $g'(x)$ and $g''(x)$, and finally, find the value of $x$ for which $g''(x) = 0$.

2. Solution Steps

First, find the first derivative $g'(x)$ using the power rule:

$\frac{d}{dx}x^n = nx^{n-1}$

$g'(x) = \frac{d}{dx}(3x^4 - 2x^2 + 49) = 3(4x^3) - 2(2x) + 0 = 12x^3 - 4x$

Next, find the second derivative $g''(x)$ by differentiating $g'(x)$ with respect to $x$:

$g''(x) = \frac{d}{dx}(12x^3 - 4x) = 12(3x^2) - 4 = 36x^2 - 4$

Now, we study the sign of $g'(x)$:

$g'(x) = 12x^3 - 4x = 4x(3x^2 - 1)$.

$g'(x) = 0$ when $4x = 0$ or $3x^2 - 1 = 0$.

This gives $x = 0$ or $x^2 = \frac{1}{3}$, so $x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3}$.

So the roots are $x = -\frac{\sqrt{3}}{3}, 0, \frac{\sqrt{3}}{3}$.

When $x < -\frac{\sqrt{3}}{3}$, $g'(x) < 0$.

When $-\frac{\sqrt{3}}{3} < x < 0$, $g'(x) > 0$.

When $0 < x < \frac{\sqrt{3}}{3}$, $g'(x) < 0$.

When $x > \frac{\sqrt{3}}{3}$, $g'(x) > 0$.

Now, we study the sign of $g''(x)$:

$g''(x) = 36x^2 - 4$.

$g''(x) = 0$ when $36x^2 - 4 = 0$.

This gives $36x^2 = 4$, so $x^2 = \frac{4}{36} = \frac{1}{9}$, so $x = \pm\frac{1}{3}$.

So the roots are $x = \pm\frac{1}{3}$.

When $x < -\frac{1}{3}$, $g''(x) > 0$.

When $-\frac{1}{3} < x < \frac{1}{3}$, $g''(x) < 0$.

When $x > \frac{1}{3}$, $g''(x) > 0$.

Finally, we solve for $x$ such that $g''(x) = 0$:

$36x^2 - 4 = 0$

$36x^2 = 4$

$x^2 = \frac{4}{36} = \frac{1}{9}$

$x = \pm\sqrt{\frac{1}{9}} = \pm\frac{1}{3}$

3. Final Answer

The values of $x$ for which $g''(x) = 0$ are $x = \frac{1}{3}$ and $x = -\frac{1}{3}$.