The problem asks us to use calculus to find the value of $x$ which minimizes the function $g(x)$. However, the function $g(x)$ is not explicitly given. From the image, a faint equation "3xdx^2 = 0" is visible. It can be assumed it's asking to find the critical point from $g'(x) = 3x+dx^2$ and that it meant to calculate $g'(x)= 3+2x$ So, let's assume that the equation is supposed to be $g'(x) = 3+2x =0$. Therefore, our goal is to find the value of $x$ for which $g'(x)=0$ using $g'(x) = 3+2x$.

1. Problem Description

The problem asks us to use calculus to find the value of $x$ which minimizes the function $g(x)$. However, the function $g(x)$ is not explicitly given. From the image, a faint equation "3xdx^2 = 0" is visible. It can be assumed it's asking to find the critical point from $g'(x) = 3x+dx^2$ and that it meant to calculate $g'(x)= 3+2x$

So, let's assume that the equation is supposed to be $g'(x) = 3+2x =0$. Therefore, our goal is to find the value of $x$ for which $g'(x)=0$ using $g'(x) = 3+2x$.

2. Solution Steps

To find the minimum of $g(x)$, we need to find the critical points of $g(x)$. We find the critical points by setting the first derivative $g'(x)$ equal to zero and solving for $x$.

$g'(x) = 3 + 2x$

Set $g'(x) = 0$:

$3 + 2x = 0$

Subtract 3 from both sides:

$2x = -3$

Divide both sides by 2:

$x = -\frac{3}{2}$

$x = -1.5$

The second derivative test is $g''(x) = 2$, Since $g''(-1.5) = 2 > 0$, we conclude that $x = -1.5$ is a minimum of g(x).

3. Final Answer

The value of $x$ that gives the minimum value of $g(x)$ is $x = -1.5$.