find [ f(x + h) - f(x) ] / h for f(x) = 100 + 2x^2
- Thread starter viberent
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Re: function defined
Really, I have been doing such problems for quite a long time. I do not try to do any of it in my head. I never have. I write down ALL intermediate steps and check all my work as I go. I always keep a large pile of scratch paper handy. You rush, you lose. Simple as that.
Very sloppy notation. It is not just nit-picky. Notation will help you if you let it. Where is the exponent?viberent said:
This is just a mess. This mess is why you managed to confuse yourself. How did 1000 get to be 100, twice? Where is the other x^2 term?
I really don't know what this means. It sounds gruesome. Why not just add or subtract? 1000 - 1000 = 0 There is no "crossing out" in there.
Nope, sorry. Try it again and BE CAREFUL. Don't just slop through it. Where did the denominator go?
Really, I have been doing such problems for quite a long time. I do not try to do any of it in my head. I never have. I write down ALL intermediate steps and check all my work as I go. I always keep a large pile of scratch paper handy. You rush, you lose. Simple as that.
Hello, viberent!
Listen carefully . . .
There are THREE steps to find the "difference quotient":
. . [1] Find \(\displaystyle f(x+h)\) . . . . . Replace \(\displaystyle x\) with \(\displaystyle x+h\) ... and simplify
. . [2] Subtract \(\displaystyle f(x)\) . . . Subtract the original function ... and simplify.
. . [3] Divide by \(\displaystyle h\) . . . . . Factor and reduce.
We are given: \(\displaystyle \:f(x)\:=\:1000 \,+\,2x^2\)
[1] Find \(\displaystyle f(x+h)\)
. . .\(\displaystyle f(x+h) \;=\;1000\,+\,2(x\,+\,h)^2 \;=\;1000\,+\,2x^2\,+\,4xh\,+\,2h^2\)
[2] Subtract \(\displaystyle f(x)\)
. . .\(\displaystyle f(x+h)\,-\,f(x)\;=\;(1000\,+\,2x^2\,+\,4xh\,+\,2h^2)\,-\,(1000\,+\,2x^2)\)
. . . . . . . . . . . . . \(\displaystyle = \;1000\,+\,2x^2\,+\,4xh\,+\,2h^2\,-\,1000\,-\,2x^2\)
. . . . . . . . . . . . . \(\displaystyle = \;4xh\,+\,2h^2\)
[3] Divide by \(\displaystyle h\)
. . .\(\displaystyle \frac{f(x+h)\,-\,f(x)}{h}\;=\;\frac{4xh\,+\,2h^2}{h} \;=\;\overbrace{\frac{2h(2x\,+\,h)}{h}}^{\text{factor}} \;=\;\overbrace{\frac{2\not{h}(2x\,+\,h)}{\not{h}}}^{\text{reduce}} \;=\;2(2x\,+\,h)\)
Listen carefully . . .
There are THREE steps to find the "difference quotient":
. . [1] Find \(\displaystyle f(x+h)\) . . . . . Replace \(\displaystyle x\) with \(\displaystyle x+h\) ... and simplify
. . [2] Subtract \(\displaystyle f(x)\) . . . Subtract the original function ... and simplify.
. . [3] Divide by \(\displaystyle h\) . . . . . Factor and reduce.
We are given: \(\displaystyle \:f(x)\:=\:1000 \,+\,2x^2\)
[1] Find \(\displaystyle f(x+h)\)
. . .\(\displaystyle f(x+h) \;=\;1000\,+\,2(x\,+\,h)^2 \;=\;1000\,+\,2x^2\,+\,4xh\,+\,2h^2\)
[2] Subtract \(\displaystyle f(x)\)
. . .\(\displaystyle f(x+h)\,-\,f(x)\;=\;(1000\,+\,2x^2\,+\,4xh\,+\,2h^2)\,-\,(1000\,+\,2x^2)\)
. . . . . . . . . . . . . \(\displaystyle = \;1000\,+\,2x^2\,+\,4xh\,+\,2h^2\,-\,1000\,-\,2x^2\)
. . . . . . . . . . . . . \(\displaystyle = \;4xh\,+\,2h^2\)
[3] Divide by \(\displaystyle h\)
. . .\(\displaystyle \frac{f(x+h)\,-\,f(x)}{h}\;=\;\frac{4xh\,+\,2h^2}{h} \;=\;\overbrace{\frac{2h(2x\,+\,h)}{h}}^{\text{factor}} \;=\;\overbrace{\frac{2\not{h}(2x\,+\,h)}{\not{h}}}^{\text{reduce}} \;=\;2(2x\,+\,h)\)