$df\over dx$|$_{x=5}$ f(x)=2 is the problem I'm working with. I believe the definition of the derivative is $\frac{f(x+h)-f(x)}{h}\,$ but how would I find the answer using this definition? There's no x variable in the problem I was given, so I'm not sure what to plug in, if anything.
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If $f(x)=2$ for all $x$ then $\frac{f(x+h)-f(x)}{h}=\frac{2-2}{h}=0$. Taking the limit again gives zero. Hence, $\frac{d}{dx} f(x)=0$ for all $x$, including $x=5$.
SpiritLevel
- 900
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You would be correct not to plug in anything, since there isn't an $x$ anywhere in the function.
So $f(x)=2$ and $f(x+h)=2$, and similarly, $f(5)=2$ and $f(5+h)=2$. Then using the definition, $$f'(5)=\displaystyle\lim_{h\to\infty}\dfrac{f(5+h)-f(5)}{h}$$ $$=\displaystyle\lim_{h\to\infty}\dfrac{2-2}{h}$$ $$\displaystyle\lim_{h\to\infty}0=0.$$
Hadi
- 402
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From the definition of $f$, $f(x+h)=2$ and $f(x)=2$. The rest is easy.