Let's solve the given problem from your photo.

The function is:

$f(x) = 10x^3 + 30x^2 - 240x + 30$

The question asks to find the values of $x$ where the derivative of $f(x)$ is zero, i.e., we need to solve $f'(x) = 0$.

Step 1: Find the derivative of $f(x)$

To find $f'(x)$, differentiate $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(10x^3) + \frac{d}{dx}(30x^2) - \frac{d}{dx} (240x) + \frac{d}{dx}(30)$

This gives:

$f'(x) = 30x^2 + 60x - 240$

Step 2: Set the derivative equal to zero

Now, we solve the equation:

$30x^2 + 60x - 240 = 0$

Step 3: Simplify the equation

Divide the entire equation by 30 to simplify:

$x^2 + 2x - 8 = 0$

Step 4: Solve the quadratic equation

Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 2$, and $c = -8$. Substituting these into the formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4 + 32}}{2}$ $x = \frac{-2 \pm \sqrt{36}}{2}$ $x = \frac{-2 \pm 6}{2}$

So, the two solutions are:

$x = \frac{-2 + 6}{2} = 2 \quad \text{or} \quad x = \frac{-2 - 6}{2} = -4$

Step 5: Final answer

The values of $x$ where the derivative is zero are:

$x = 2, -4$

Would you like more details or have any questions?

Related questions:

1. How can we verify if $x = 2$ and $x = -4$ are indeed the critical points?
2. What is the significance of the second derivative in determining whether these points are maxima, minima, or points of inflection?
3. How do we solve higher-order polynomial derivatives?
4. Could you explain how the quadratic formula works in different cases, such as when the discriminant is negative?
5. How do we factor quadratic equations without using the quadratic formula?

Tip: Always simplify equations before applying methods like the quadratic formula for easier calculations!