Question

How small can $f(4)$ possibly be? If $f$ is a function such that $f(1){\text{ }} = {\text{ }}4$ and $f'(x){\text{ }} \geqslant {\text{ }}2$ for $1{\text{ }} \leqslant {\text{ }}x{\text{ }} \leqslant {\text{ }}4$ respectively.
(a) $11$
(b) $16$
(c) $10$
(d) $15$

Answer 1

Hint : The given problem revolves around the concepts “mean value theorem”, which states that any function say, $f$ is is continuous on $\left[ {a,b} \right]$ and differentiable on same $\left[ {a,b} \right]$ then it exists $f'\left( x \right){\text{ }} = {\text{ }}\dfrac{{f\left( b \right){\text{ }} - {\text{ }}f\left( a \right)}}{{b{\text{ }} - {\text{ }}a}}$ respectively. As a result, substituting the given values in this formula the desired value is obtained.

Complete step-by-step answer :
Since, we have given that
‘$f$’ is a function for the equations $f'(x){\text{ }} \geqslant {\text{ }}2$ at $1 \leqslant x \leqslant 4$ and $f(1){\text{ }} = {\text{ }}4$ respectively,
As a result,
Let us assume that ‘$f\left( x \right)$’ be the function satisfying the following hypothesis that is;
(i) $f\left( x \right)$ is continuous on a closed interval, that is $f\left( 1 \right){\text{ }} = {\text{ }}4$.
(ii) $f'\left( x \right)$ is differentiable on open interval $\left[ {a,b} \right]$ that is $\left[ {1,4} \right]$
Therefore, $f\left( x \right)$ is continuous on $\left[ {1,4} \right]$ respectively!
Hence, applying the Mean Value Theorem $f'\left( x \right){\text{ }} = {\text{ }}\dfrac{{f\left( b \right){\text{ }} - {\text{ }}f\left( a \right)}}{{b{\text{ }} - {\text{ }}a}}$, we get
Substituting the given boundary conditions tends to the function $f$ that is $a \leqslant x \leqslant b{\text{ }} \to {\text{ }}1 \leqslant x \leqslant 4$ i.e. $b = 4$ and $a = 1$, we get
$f'\left( x \right){\text{ }} = {\text{ }}\dfrac{{f\left( 4 \right){\text{ }} - {\text{ }}f\left( 1 \right)}}{{4{\text{ }} - {\text{ }}1}}$
$f'\left( x \right){\text{ }} = {\text{ }}\dfrac{{f\left( 4 \right){\text{ }} - {\text{ }}f\left( 1 \right)}}{3}$
But, we have given that $f\left( 1 \right) = 4$,
So, substituting $f\left( 1 \right) = 4$ in the equation, we get
$3f'\left( x \right){\text{ }} = {\text{ }}f\left( 4 \right){\text{ }} - {\text{ }}4$
$f\left( 4 \right){\text{ }} = {\text{ }}4{\text{ }} + {\text{ }}3{\text{ }}f'\left( x \right)$
We have given that $f'(x) \geqslant 2$, we get
$f\left( 4 \right){\text{ }} = {\text{ }}4{\text{ }} + {\text{ }}3\left( 4 \right)$
$f\left( 4 \right){\text{ }} = {\text{ }}4{\text{ }} + {\text{ }}12$
Hence, the required solution reveals that,
$\therefore \Rightarrow$ $f\left( 4 \right){\text{ }} = {\text{ }}16$
Since, $f'(x){\text{ }} \geqslant {\text{ }}2$ contains at least $2$ value,
Hence, $f(4)$ satisfies the value at least $16$ i.e. $f\left( 4 \right){\text{ }} \geqslant {\text{ }}16$.
$\Rightarrow \therefore$The option (b) is correct.
So, the correct answer is “Option b”.

Note : One must be able to know the Mean Value Theorem with given conditions that are equations to find the respective condition or any value for the function provided to us. Consideration of boundary conditions should be analyse properly with respect to the Mean value formula $f'\left( x \right){\text{ }} = {\text{ }}\dfrac{{f\left( b \right){\text{ }} - {\text{ }}f\left( a \right)}}{{b{\text{ }} - {\text{ }}a}}$ and solve the equation mathematically, so as to be sure of our final answer.