g(x)=2x^2-3x+10; g'(x)=4x-3
How would you find g'(x) when g(x) =15?

Question

g(x)=2x^2-3x+10; g'(x)=4x-3
How would you find g'(x) when g(x) =15?

anonymous · Answer

didn't you already find g'(x)?

anonymous · Answer

Or if you mean g'(15), You would find the derivative of $g(x)$ and evaluate this at 15. Since you have most genourously already obtained the derivative, just evaluate it haha:
$$\eqalign{
&g'(x)=4x-3 \
&g'(15)=4(15)-3 \
&g'(15)=60-3\
&g'(15)=57 \
}$$

anonymous · Answer

This was how the question was stated. It seemed weird to me.

ddcamp · Answer

Solve g(x) = 15 for x
2x² - 3x + 10 = 15 → 0 = 2x² - 3x - 5 → x= 2.5, -1

Then, plug those values of x into g'(x)

campbell_st · Answer

substitute g(x) with 15

and you get

$$15 = 2x^2 - 3x + 10$$

solve for x

when you get x, subsitute into g'(x)

anonymous · Answer

Haha that was my third guess

anonymous · Answer

Well is g(x)=15, the derivative is really easy to calculate. The derivative of a constant is 0. So g'(x)=0.