Let $f(x)$ be a differentiable function in $R\rightarrow R$ then how to obtain the derivative of $g(x)$ based on the following information

Question

Let $f:R\rightarrow R$ be a differentiable function such that $f(2)=2$ and

$\left | f(x)-f(y) \right |\leq5\left ( \left | x-y \right | \right )^{\frac{3}{2}}$, and $g(x)=x^3f(x)$, then what is the value of $g'(2)$.

Since, $g(x)=x^3f(x)$, by taking derivative both sides, we get:

$g'(x)=x^3f'(x)+3x^2f(x)$ which gives $g'(2)=8f'(2)+24$ on using the information $f(2)=2$. The answer is $24$, so I am sure that the $f'(2)$ will vanish but not able to understand how. If anybody could help me with the problem, it will be very beneficial for me. Probably, I have to use the $\left | f(x)-f(y) \right |\leq5\left ( \left | x-y \right | \right )^{\frac{3}{2}}$, but I am not sure how can I use this?.

Answer 1

\begin{align*} \dfrac{\left|f(x)-f(2)\right|}{|x-2|}\leq 5|x-2|^{1/2}\rightarrow 0, \end{align*} as $x\rightarrow 2$.

Answer 2

Use the fact $$\left | f(x)-f(y) \right |\leq5\left ( \left | x-y \right | \right )^{\frac{3}{2}}$$ to show $f'(2)=0.$

Note that $$f'(2) =lim_{X\to 2 \frac {f(x)-f(2)}{x-2}} $$

$$|f(x)-f(2)|\le 5|x-2|^{3/2}$$

Therefore $$|\frac {f(x)-f(2)}{x-2}| \le 5|x-2|^{1/2} $$

Note that $$ lim_{x\to 0} 5|x-2|^{1/2} = 0$$ Thus $f'(2)=0$