# Find the average rate of change of $2x^3 - 5x$ on the interval $[1,3]$

Find the average rate of change of $2x^3 - 5x$ on the interval $[1,3]$.

I'm really confused about this problem. I keep ending up with the answer $12$, but the answer key says otherwise. Someone please help! Thanks!

• How did you get 12? – Aaron Maroja Sep 11 '15 at 18:32
• add the answer from the book and write what you have tried – Iuli Sep 11 '15 at 19:27

The total change is $h(3)-h(1)=52$. The length of the interval is $2$. So the average rate of change is $52/2=26$.
Update: You have apparently changed the function in the question from $h(x)=2x^3-5$ to $h(x)=2x^3-5x$. The answer for the new function is $\frac {h(3)-h(1)}2=21$.
average rate of change is the difference between the function values at two points divided by the distance between the points. $$= \frac{f(3) - f(1) }{3-1}$$ $$= \frac{[2*(3^3)-5*(3)] -[2*(1^3)-5*(1)]}{2}$$ $$= \frac{[54-15]-[2-5] }{2}$$ $$= \frac{39-(-3) }{2}$$ $$= \frac{42 }{2}$$ $$= \frac{21} 1$$