Question

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) ) .

Answer 1

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

Answer 2

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

Answer 3

ohh

Answer 4

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

Answer 5

Think you can handle the rest?

Answer 6

i got, thank you!

Answer 7

sweet.

Answer 8

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

Answer 9

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

Answer 10

ohhh thats very helpful, thanks

Answer 11

mmhmm, no problem

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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.

Answer 17

how would i do c?

Answer 18

so you have (x, f(x) ) and (x+h, f(x+h) )

Which translates into \[m= \frac{f(x+h)-f(x)}{(x+h)-x}\]

Answer 19

oh i just leave the f(x)

Answer 20

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