Question

$\rm The \: table \: below \: lists \: some \: values \: for \: the \\ \rm differentiable \: function \: h \: and \: its \: first \\ \rm derivative \: h'. \\ \begin{gathered}\begin{array}{ |c| c| c| c | c| c| c| } \hline \rm x&2&6&7&9&11&13 \\ \hline \rm h(x)&15& - 9& - 10& - 6&6&26 \\ \hline \rm h'(x)& - 10& - 2&0&4&8&12 \\ \hline\end{array}\end{gathered} \\ \rm Define \: f(x) = \int_{0}^{x} h(t)dt. \\ \rm What \: is \: the \: least \: possible \: number \: of \: \\ \rm critical \: points \: of \: f \: in \: the \: interval \\ \rm 2 \le x \le 13 \: ?$​
A) Zero
B) One
C) Two
D) It cannot be determined from the information given.

Answer 1

$\large\underline{\sf{Solution-}}$

Given information is

$\\ \begin{gathered}\begin{array}{ |c| c| c| c | c| c| c| } \hline \rm x&2&6&7&9&11&13 \\ \hline \rm h(x)&15& - 9& - 10& - 6&6&26 \\ \hline \rm h'(x)& - 10& - 2&0&4&8&12 \\ \hline\end{array}\end{gathered}$

Further given that,

$\rm \: f(x) = \displaystyle\int_{0}^{x} \rm h(t)dt$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}f(x) = \dfrac{d}{dx}\displaystyle\int_{0}^{x} \rm h(t)dt$

Now, on differentiating using Leibnitz Rule, we get

$\rm \: f'(x) = \displaystyle\int_{0}^{x} \rm \: \bigg(\frac{\partial }{\partial x} h(t)\bigg)dt + \dfrac{d}{dx}(x) \: [h(x)] - \dfrac{d}{dx}(0)$

$\rm \: f'(x) = 0 + 1 \times \: [h(x)] - 0$

$\rm \: f'(x) = h(x)$

For critical points,

$\rm \: f'(x) = 0$

$\rm\implies \:h(x) = 0$

But from given data, we concluded that

$\rm \: h(x) \: \ne \: 0 \:for \:few\:values\:of \: x \: \in \: [2,13]$

But its not concluded from the given data that

$\rm \: h(x) \: \ne \: 0 \: \forall \: x \: \in \: [2,13]$

So, its not possible to find critical points from the given information.

So, option (D) is correct.