Let f:RR be a continuously differentiable function such that f(2)=6 and f'(2)=148.

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be a continuously differentiable function such that $f(2)=6$ and $f'(2)=\dfrac1{48}$. If $\displaystyle\int_6^{f(x)}4t^3\,dt=(x-2)g(x)$, then $\displaystyle\lim_{x	o2}g(x)$ is equal to:

Accepted Answer

Differentiating both sides of $\displaystyle\int_6^{f(x)}4t^3\,dt=(x-2)g(x)$ with respect to $x$:
$4(f(x))^3f'(x)=g'(x)(x-2)+g(x)$

Setting $x=2$:
$$4(f(2))^3f'(2)=g(2)\Rightarrow4(6)^3\cdot\frac1{48}=g(2)\Rightarrow\lim_{x\to2}g(x)=g(2)=\frac{4\cdot216}{48}=18$$

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