# Average rate of change?

How would I figure the following problem out?

Find the average rate of change of $g(x)=x^2+3x+7$ from $x=5$ to $x=9$

My thought is that I would plug in 5 and 9 for the x values to get the y values. And the use the slope formula $\frac{y_2-y_1}{x_2-x_1}$.

• Sounds good! The average rate of change is the net change, divided by "how long" it took to make that change. – André Nicolas Feb 5 '13 at 2:14
• I got the average rate of change as 17! – Fernando Martinez Feb 5 '13 at 2:17
• It is indeed $17$! – David Mitra Feb 5 '13 at 2:20
• Nice job, Fernando! – Namaste Feb 5 '13 at 2:21
• @FernandoMartinez: Note you can do it by $(a^2+3a+7)-(b^2+3b+7)=(a-b)(a+b)+3(a-b)$. Divide by $a-b$, you get $a+b+3$. Or one can calculate. – André Nicolas Feb 5 '13 at 2:26

The good thing here is that you have a quadratic equation, so differentiating w.r.t. $x$ , you get a linear equation...which makes it easy to find average rate of change.

Consider your equation $$g(x)=x^2+3x+7$$ differentiate it w.r.t. $x$ $$g'(x)=2x+3$$

The naive way of doing this question is finding $g'(5), g'(6),g'(7), g'(8),g'(9)$ and then finding its average. Alternatively, just find the average of $g'(5)$ and $g'(9)$. You will get 17.

• But do I have to use calculus to solve it or is finding the slope not good enought. – Fernando Martinez Feb 5 '13 at 2:19
• Differentiating it w.r.t. $x$ is the "using calculus" part. – bryanblackbee Feb 5 '13 at 2:33
• @FernandoMartinez: Very insightful question in your comment above. The answer to your question is "yes and no". "No" because if you drive a total distance of 10 miles in 30 minutes, your average speed will be 20 miles per hour regardless of what your speedometer read during the drive. "Yes" because that is the Fundamental Theorem of Calculus. – Barbara Osofsky Feb 5 '13 at 15:06

Yes, your solution is correct.

Differentiating first is a detour.

The Fundamental Theorem of Calculus says that $\int_a^b f(t)\ dt=F(b)-F(a)$ where the rate of change of $F(t)$ with respect to $t$ is $f(t)$. The only thing that matters is where you start and where you end. Slight observation: if you have to make sure if you apply this to something like average speed as long as you used distance traveled rather than distance from start to end because speed is always a non-negative absolute value of velocity.

In the case of this question, this says the average rate of change of $g(x)$ (as opposed to the average change of $g$), is $$\mathrm{Average\ rate\ of\ change} ={{g(b)-g(a)} \over {b-a}}={ {g(9)-g(5)}\over {9-5}}= {{9^2+3\cdot 9+7)-(5^2+3\cdot 5+7)}\over {9-5}}={{115-47}\over {47}}=17$$ and this is exactly what a precalculus student should do.