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A function is given. Determine the average rate of change of the function between the given values of the variable.

$f(x) = 2 − x^2 $

$x = 8, x = 8 + h$

I solved for $f(8)$ and got $-62$...

I need help solving for $f(8+h)$. I get down to $2-(h^2+16h+64)$ and I don't know what to do after that; no matter what I try I get stuck. I know how to do the rest after that (plug it into the formula: change in y / change in x and simplify).

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  • $\begingroup$ $2 - (h^2 + 16h + 64) = 2 - h^2 - 16h - 64 = -62 -16h - h^2$. $\endgroup$ – littleO Jul 3 '15 at 1:15
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$$AROC=\frac{f(b)-f(a)}{b-a}$$ $$=\frac{f(8+h)-f(8)}{(8+h)-8}$$ $$=\frac{[2-(8+h)^2]-[2-8^2]}{(8+h)-8}$$ $$=\frac{[2-(64+16h+h^2)]-[2-64]}{(8+h)-8}$$ $$=\frac{(2-64-16h-h^2)-(2-64)}{(8+h)-8}$$ $$=\frac{2-64-16h-h^2-2+64}{8+h-8}$$ $$=\frac{-16h-h^2}{h}$$ $$=\frac{h(-16-h)}{h}$$ $$=-16-h$$

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  • $\begingroup$ Thanks, I didn't know you could do it all at once. I was solving them seperatly. $\endgroup$ – TheNewGuy Jul 3 '15 at 1:11
  • $\begingroup$ You can solve them separately, but I teach my class to do it this way, to keep sight of the overall picture. This way also leads better to some related ideas in calculus. That's why I did not combine the $2-64$: you'll see in calculus. $\endgroup$ – Rory Daulton Jul 3 '15 at 1:12
  • $\begingroup$ Yeah seeing the bigger picture helps a lot. Thank you again! $\endgroup$ – TheNewGuy Jul 3 '15 at 1:14
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    $\begingroup$ Much quicker than expanding the binomial square is to just note that the $2$s cancel out and the expression on the numerator becomes $8^2 - (8+h)^2 = (16+h)(-h)$ after application of $a^2 - b^2 = (a+b)(a-b)$. In fact, I did it mentally (and quickly) this way. :) $\endgroup$ – Deepak Jul 3 '15 at 1:32

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