# Find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$

I need to find the mean slope of $$f(x) = 2x^3-6x^2-90x+6$$ on the interval $$[-5,8]$$. I am getting $$\frac{518}{13}$$ but this is wrong. Here is my work.

Using the mean value theorem $$f'(c)=\frac{f(b) - f(a)}{b - a}$$

$$f(b) = 2(-5)^3-6(-5)^2-90(-5)+6 = -250-150+450+6 = 56$$

$$f(a) = 2(8)^3-6(8)^2-90(8)+6 = 1024-384-66 = 574$$

$$\frac{56-574}{-5-8}=\frac{518}{13}$$

• The mean slope is just the slope of the secant line. – coreyman317 Mar 31 at 20:57

Your error is a simple mistake in computing $$f(8)$$. To be precise, you write that $$2(8)^3-6(8)^2-90(8)+6 = 1024-384-66,$$ where indeed $$2\times8^3=1024$$ and $$6\times8^2=384$$. You seem to have computed $$9\times8$$ instead of $$90\times8$$.
• Yes, but you compute $9\times8$ instead of $90\times8$ (I have updated my answer to point this out more clearly) – Servaes Mar 31 at 21:02