are these excercises right?

G

georgebaseball

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Sep 4, 2006
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Hello

they give me the equation x^2-2 and ask me to find f(x+h)-f(x)/h
and I got as answer 2x+h-2, is that right?

and they ask me to find the average rate of change from x=1 to x=2 in the equation x-2x^2
in this one I got 0 as answer

thanks
 
Hello

they give me the equation x^2-2 and ask me to find f(x+h)-f(x)/h
and I got as answer 2x+h-2, is that right?
Click to expand...

You're slightly erring.

\(\displaystyle \L\\\frac{f(x+h)-f(x)}{h}\)

You have:

\(\displaystyle \L\\\frac{(x-h)^{2}-2-(x^{2}-2)}{h}\)

\(\displaystyle \L\\\frac{x^{2}+2hx+h^{2}-2-x^{2}+2}{h}\)

\(\displaystyle \L\\\frac{2hx+h^{2}}{h}\)

\(\displaystyle \L\\2x+h\)

\(\displaystyle \L\\\lim_{h\to\0}2x+h=2x\)
 
No, the average rate of change is:

\(\displaystyle \L\\m_{sec}=\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\)

\(\displaystyle x_{0}=1 \;\ amd \;\ x_{1}=2\)
 
Hello, georgebaseball!

Sorry, both your answers are wrong.

Galactus went one step too far (an understandable error) . . .

Given: \(\displaystyle f(x) \:=\:x^2\,-\,2.\;\;\)Find \(\displaystyle \,\frac{f(x+h)\,-\,f(x)}{h}\)
Click to expand...

That "difference quotient" is comprised of three steps:
. . (1) Find \(\displaystyle f(x+h)\) ... and simplify
. . (2) Subtract \(\displaystyle f(x)\) ... and simplify
. . (3) Divide by \(\displaystyle h\) ... and simplify

(1) \(\displaystyle f(x+h)\;=\;(x+h)^2\,-\,2\;=\;x^2\,+\,2xh\,+\,h^2\,-\,2\)

(2) \(\displaystyle f(x+h)\,-\,f(x)\;=\;(x^2\,+\,2xh\,+\,h^2)\,-\,(x^2\,-\,2)\;=\;2xh\,+\,h^2\)

(3) \(\displaystyle \L\frac{f(x+h)\,-\,f(x)}{h}\;=\;\frac{2xh\,+\,h^2}{h}\;=\;\frac{\not{h}(2x\,+\,h)}{\not{h}}\)\(\displaystyle \;=\;\fbox{2x\,+\,h}\)


Find the average rate of change of \(\displaystyle \,f(x)\:=\:x\,-\,2x^2\,\) from \(\displaystyle x\,=\,1\) to \(\displaystyle x\,=\,2.\)
Click to expand...

\(\displaystyle \text{average rate of change of }f(x)\;=\;\frac{\text{change in }f(x)}{\text{change in }x}\)


When \(\displaystyle x\,=\,1:\;f(1)\:=\:1\,-\,2\cdot1^2\:=\:-1\)
When \(\displaystyle x\,=\,2:\;f(2)\:=\:2\,-\,2\cdot2^2\:=\:-6\)
. . The change in \(\displaystyle f(x)\) is: \(\displaystyle \,(-6)\,-\,(-1)\;=\;-5\)

The change in \(\displaystyle x\) is: \(\displaystyle \,2\,-\,1\;=\;1\)


Therefore: \(\displaystyle \:\text{average rate of change }\,=\;\frac{-5}{1} \;=\;\fbox{-5}\)

 
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