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let $f$ be a differentiable function and $$\lim_{x\to 4}\dfrac{f(x)+7}{x-4}=\dfrac{-3}{2}.$$

Define $g(x)=\dfrac{f(2x)}{x}$. I want to know the derivative $$\dfrac{d}{dx}g(2)=?$$


I know that :

$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$

and :

$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}=a\in\mathbb{R}$$

so :

$$\lim_{x\to 4}\dfrac{f(x)-f(4)+7+f(4)}{x-4}=\dfrac{-3}{2}$$

$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}+\dfrac{f(4)+7}{x-4}=\dfrac{-3}{2}$$

now what ?

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  • 1
    $\begingroup$ $\frac { dg\left( 2 \right) }{ dx } =0,$ perhaps you mean derivative $g(x)$at the point $x=2$? $\endgroup$
    – haqnatural
    Jul 8, 2017 at 15:09
  • $\begingroup$ $g'(2)\ne \frac{4f'(4)-\color{red}{4}f(4)}{4}$. Rather, $$g'(2)=\frac{4f'(4)-f(4)}{4}$$ $\endgroup$
    – Mark Viola
    Jul 8, 2017 at 15:13

4 Answers 4

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You have an extra $4$ in the numerator here:

i know that :

$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$

If $g(x) = \dfrac{f(2x)}x$, then

\begin{align*} \frac d{dx} g(x) &= \frac d{dx} \left(\frac{f(2x)}x\right)\\[0.3cm] &= \frac{x \frac d{dx} f(2x) - f(2x) \frac d{dx}x}{x^2}\\[0.3cm] &= \frac{2x f'(2x) - f(2x)}{x^2}\\[0.3cm] \end{align*}

By $\dfrac d{dx} g(2)$ I think you mean $\dfrac d{dx} g(x)\bigg|_{x=2}$. Or, more compactly, $g'(2)$.

So $g'(2) = \dfrac{4f'(4) - f(4)}4$.

We're given $\displaystyle \lim_{x\to 4} \frac{f(x) + 7}{x-4} = -\frac32$. Since $\lim_{x\to4} (x-4) = 4-4 = 0$, then $\displaystyle\lim_{x\to4} (f(x) + 7)$ must also be zero because this is the only way the limit of the quotient can be a finite number. Also, since $f$ is continuous (because it is differentiable) then we can safely say $\displaystyle\lim_{x\to 4}(f(x) + 7) = f(4) + 7$. Thus, $f(4) + 7 = 0$, and so we have $f(4) = -7$.

It remains to find $f'(4)$. By the definition of the derivative (sometimes called the alternate definition of the derivative), we have: $$ f'(4) = \lim_{x \to 4} \frac{f(x) - f(4)}{x-4} = \lim_{x \to 4} \frac{f(x) + 7}{x - 4}$$ And we were given that this is $-\dfrac32$.

Now plug all the numbers into where they need to go, and simplify.

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Since $f$ is differentiable with

$$f'(4)=\lim_{x\to 4}\frac{f(x)-(-7)}{x-4}=-\frac32$$

Then, $f(4)=-7$.

Now, $g(x)=\frac{f(2x)}{x}\implies g'(x)=\frac{2xf'(2x)-f(2x)}{x^2}$, from which we have

$$g'(2)=\frac{4f'(4)-f(4)}{4}=\frac{-6+7}{4}=\frac14$$

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You then got:

$$f'(4)+\lim_{x\to 4}\frac{f(4)+7}{x-4}=-\frac{3}{2}$$

so you must have $f(4)=-7$ because, if not, you will have $\frac{f(4)+7}{x-4}\to \pm\infty$ and it is impossible. Futhermore you get $f'(4)=-3/2$. Now just use it in your expression of $g'(2)$

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HINT

\begin{align*} \frac{dg}{dx}(2) & = \lim_{x\rightarrow 2} = \frac{g(x) - g(2)}{x - 2} = \lim_{x\rightarrow 2}\frac{\displaystyle\frac{f(2x)}{x} - \frac{f(4)}{2}}{x - 2} = \lim_{x\rightarrow 2}\frac{2f(2x)-xf(4)}{2x(x-2)}\\ & = \lim_{u\rightarrow 4}\frac{2f(u) - u\displaystyle\frac{f(4)}{2}}{u\left(\displaystyle\frac{u}{2}-2\right)} = \lim_{u\rightarrow 4}\frac{4f(u)-uf(4)}{u(u-4)} = \lim_{u\rightarrow 4}\frac{4f(u)-uf(4)}{u^2 - 4u}\\ & \overset{\mathrm{L'H}}{=}\lim_{u\rightarrow 4}\frac{4f'(u)-f(4)}{2u-4} = \frac{4f'(4)-f(4)}{4} \end{align*}

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