aphathalo

2022-09-07

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=9My 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 $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

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=9My 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 $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

Elliott Rollins

Beginner2022-09-08Added 8 answers

The Fundamental Theorem of Calculus says that ${\int}_{a}^{b}f(t)\text{}dt=F(b)-F(a)$ 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$\mathrm{A}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\text{}\mathrm{o}\mathrm{f}\text{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}=\frac{g(b)-g(a)}{b-a}=\frac{g(9)-g(5)}{9-5}=\frac{{9}^{2}+3\cdot 9+7)-({5}^{2}+3\cdot 5+7)}{9-5}=\frac{115-47}{47}=17$and this is exactly what a precalculus student should do.

Gardiolo0j

Beginner2022-09-09Added 1 answers

Yes, your solution is correct.

Differentiating first is a detour.

Differentiating first is a detour.

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