The problem asks to estimate the interval(s) on which the rate of change of the function $f(x) = 10x^3 + 3x^2 - 12x$ will be positive, based on the given graph of the function. The rate of change of a function is positive when the function is increasing.

1. Problem Description

The problem asks to estimate the interval(s) on which the rate of change of the function $f(x) = 10x^3 + 3x^2 - 12x$ will be positive, based on the given graph of the function. The rate of change of a function is positive when the function is increasing.

2. Solution Steps

The rate of change is positive when the function is increasing.

From the graph, we can see that the function is increasing when $x < -0.75$ and when $x > 0.55$.

However, we need to determine which answer choice represents this interval or a subinterval of these regions.

Looking at the provided options:

- $x < -0.75, x > 0.55$ : This seems to be the best answer as the intervals where the function is increasing are approximately $x < -0.75$ and $x > 0.55$

- $0 < x < 0.95$ : The function decreases then increases on this interval.

- $-1.5 < x < 0$ : The function decreases then increases on this interval.

- $-0.75 < x < 0.55$ : The function decreases on this interval.

To confirm that the rate of change is positive when $x < -0.75$ and $x > 0.55$, we can find the derivative of $f(x)$:

$f'(x) = 30x^2 + 6x - 12$.

We want to find when $f'(x) > 0$.

$30x^2 + 6x - 12 > 0$

$5x^2 + x - 2 > 0$

We can solve $5x^2 + x - 2 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4(5)(-2)}}{10} = \frac{-1 \pm \sqrt{1 + 40}}{10} = \frac{-1 \pm \sqrt{41}}{10}$

$x_1 = \frac{-1 - \sqrt{41}}{10} \approx \frac{-1 - 6.4}{10} \approx -0.74$

$x_2 = \frac{-1 + \sqrt{41}}{10} \approx \frac{-1 + 6.4}{10} \approx 0.54$

Therefore $f'(x) > 0$ when $x < -0.74$ or $x > 0.54$.

From the options, the interval $x < -0.75, x > 0.55$ is the correct one.

3. Final Answer

$x < -0.75, x > 0.55$