Consider the function $f(x) = 2 x^3 - 9 x^2 - 24 x+ 2$ on the interval $[ -5 , 10 ]$. Find the average or mean slope of the function on this interval.
Average slope = $81$ (Is Correct)
By the Mean Value Theorem, we know there exists at least one $c$ in the open interval $( -5 , 10 )$ such that $f'( c)$ is equal to this mean slope. Find all values of $c$ that work and list them.
$$ 6x^2-18x-57$$
$$ \frac{18\pm\sqrt{(-18)^2-4(6)(-57)}}{12} $$
$$ \frac{18-\sqrt{1692}}{12} =-1.9278273 $$ $$ \frac{18+\sqrt{1692}}{12} = 4.9278273 $$
Then,
$$ c = -1.9278273, 4.9278273 $$
Correct Answer: $$ c = -2.94409720865779,5.94409720865779 $$
I don't know what method they used to get those numbers, or if my calculation has an error.
$$6x^2-18x-24=81$$
$$6x^2-18x-24-81=0 (\text{I believe your mistake is at this step})$$
$$6x^2-18x-105=0$$ Can you solve the following quadratic equation? $$2x^2-6x-35=0$$