Question

URGENT
Let f and g be twice differentiable functions such that fog(x) = x^3 and g(1) = 2, g'(1) = 1, g" (1) = 4 then f" (2) is equal to
A) 0
B) 2
c) 4
d) -6​

Answer 1

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

Given that,

Let f and g be twice differentiable functions such that

• g(1) = 2

• g'(1) = 1

• g" (1) = 4

and

$\rm :\longmapsto\:fog(x) = {x}^{3}$

$\rm :\longmapsto\:f[g(x)] = {x}^{3}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} f[g(x)] = \dfrac{d}{dx} {x}^{3}$

$\rm :\longmapsto\:f'[g(x)]\dfrac{d}{dx} g(x) = {3x}^{3 - 1}$

$\rm :\longmapsto\:f'[g(x)] \: g'(x) = {3x}^{2}$

On substituting x = 1, we get

$\rm :\longmapsto\:f'[g(1)] \: g'(1) = 3$

$\rm :\longmapsto\:f'[2] \: \times 1 = 3$

$\rm :\longmapsto\:\boxed{ \tt{ \: f'[2] = 3 \: }} - - - - (1)$

Now, Again, from above we have

$\rm :\longmapsto\:f'[g(x)] \: g'(x) = {3x}^{2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} f'[g(x)] \: g'(x) = \dfrac{d}{dx} {3x}^{2}$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} u.v \: = \: u \: \dfrac{d}{dx} v \: + \: v \: \dfrac{d}{dx} u \: }}$

So, using this, we get

$\rm :\longmapsto\:f'[g(x)]\dfrac{d}{dx} g'(x) + g'(x)\dfrac{d}{dx} f'[g(x)] = 6x$

$\rm :\longmapsto\:f'[g(x)] g''(x) + g'(x) f''[g(x)] g'(x)= 6x$

On substituting x = 1, we get

$\rm :\longmapsto\:f'[g(1)] g''(1) + g'(1) f''[g(1)] g'(1)= 6$

$\rm :\longmapsto\:f'[2] \times (4) + 1 \times f''[2] \times 1= 6$

$\rm :\longmapsto\:3 \times 4 + f''[2] = 6$

$\rm :\longmapsto\:12 + f''[2] = 6$

$\rm :\longmapsto\: f''[2] = 6 - 12$

$\rm :\longmapsto\: \boxed{ \tt{ \: f''[2] \: = - \: 6 \: }}$

• Hence, Option (d) is correct

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$