It is easier to take the derivative if you first expand $(x+2)^2(x-5)$ into a standard cubic $ax^3 + bx^2 + cx + d$. Try multiplying it out.

After you compute $f'(x)$, set it equal to 0 and solve for $x$. You will get two $x$-values; think about the general shape of a cubic with positive leading coefficient to decide which one is the local minimum.

In Desmos, type f(x) = (x+2)^2(x-5) and make sure the viewing window shows roughly from $x=-6$ to $x=6$ and enough $y$-values to see the top and bottom of the curve.

Look at the graph: you should see the curve touch the $x$-axis at $x=-2$ (a local high point) and then dip down to a low point before rising again. Zoom in around that "valley" to the right of $x=-2$ and tap or click on the lowest point of the curve to see its coordinates.

Note the $x$-coordinate of that lowest point (the local minimum). Compare this $x$-value with the answer intervals and select the option whose interval contains that value.

Start by expanding $(x+2)^2(x-5)$ so it is easier to differentiate.

First compute $(x+2)^2$:

Now multiply by $(x-5)$:

So we can write

Local maxima and minima occur where the slope of the curve (the derivative) is zero.

Differentiate term by term:

The $x$-coordinates of local extrema are solutions of

Solve $3x^2 - 2x - 16 = 0$.

This quadratic factors as

Set each factor equal to zero:

• $3x - 8 = 0 \Rightarrow x = \frac{8}{3}$
• $x + 2 = 0 \Rightarrow x = -2$

So the graph has two critical points at $x = -2$ and $x = \frac{8}{3}$.

For a cubic with positive leading coefficient (like $x^3 - x^2 -16x -20$), the graph goes down to $-\infty$ on the left and up to $+\infty$ on the right. That means:

• The left critical point is a local maximum.
• The right critical point is a local minimum.

Here, $x=-2$ is left of $x=\frac{8}{3}$, so $x=-2$ is the local maximum and $x=\frac{8}{3}$ is the local minimum.

Now estimate $\frac{8}{3} \approx 2.67$, which lies between 2 and 3. Therefore, the $x$-coordinate of the local minimum is in the interval $2 < x < 3$.