anonymous one year ago Consider the function f(x) = -5 x - 2 and find the following: a) The average rate of change between the points (-1, f(-1) ) and (1, f( 1 ) ) . b) The average rate of change between the points (a, f(a) ) and (b, f(b) ) . c) The average rate of change between the points (x, f(x) ) and (x+h, f(x+h) ) .

1. anonymous

Average rate of change : $$A(x) = \dfrac{f(x)-f(a)}{x-a}$$

2. anonymous

where do i find a if like in a) i plug in -1 to the f(x) which is -5(-1)-2

3. anonymous

ohh

4. anonymous

$A(x) = \frac{f(-1) - f(1)}{-1 -(1)}$

5. anonymous

Think you can handle the rest?

6. anonymous

i got, thank you!

7. anonymous

sweet.

8. anonymous

Just remember that the 2 points represent 2 coordinate points, and you're trying to find the slope $A(x) = \frac{f(x)-f(a)}{x-a} \iff m = \frac{y_2 -y_1}{x_2-x_1}$

9. anonymous

When you think about it this way, it's easier to identify which points subtract what, I guess

10. anonymous

11. anonymous

mmhmm, no problem

12. anonymous

wait can you check my answers cause for a, i got -1 and for b i got a^2-b^2-5. I think i plugged in wrong

13. anonymous

Think I might have stated the wrong values of x, but this is how i solved it. $f(a)=f(-1) = 3~,~f(x)= f(1) = -7$$m = \frac{f(x)-f(a)}{x-a} = \frac{-7-(3)}{1-(-1)} =-\frac{10}{2}=-5$

14. anonymous

oh wow okay i was looking at a similar but different problem...I think i can do it now, thanks!

15. Astrophysics

Yeah it's basically $\frac{ y_2-y_1 }{ x_2-x_1 }$ notice it's the slope formula @Jhannybean it's better to write the equation as $m = \frac{ f(b)-f(a) }{ b-a }$ to avoid confusion

16. anonymous

And i figured I switched it because when you solve the last one with $$(x~,~ f(x))$$ and $$(x+h,~f(x+h))$$ you will get the definition of a derivative once you take the limit of it, but that's for later i guess. Therefore I knew i made a mistake.

17. anonymous

how would i do c?

18. anonymous

so you have (x, f(x) ) and (x+h, f(x+h) ) Which translates into $m= \frac{f(x+h)-f(x)}{(x+h)-x}$

19. anonymous

oh i just leave the f(x)

20. anonymous

Hm? Nothing changes in the numerator, it's only the denominator that you're concerned wtih.