Question

Let f(x) be a differentiable function and g(x) = (f(x))^3-3(f(x))^2+4f(x) +3sinx+4cosx+5x for all x belongs to real number, prove that g is always increasing whenever f is increasing. ​

Answer 1

g(x)=f(x)3−3f(x)2+4f(x)+5x+3sinx+4cosx  

g(x)3f(x)2−6f(x)+4f(x)+5+3cosx−4sinx

Now 3f(x)2−6f(x)+4=3f(x)−12+1l>0

and -5 ≤3cosx−4sinx≤5

∴0≥3cosx−4sindx+5≤10

When f(x) increases f(x) ≥0

So from (1),g(x)≥0

So ,g(x) in increasing whenever f(x) increases.

Answer 2

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

Given that,

$\rm :\longmapsto\:f(x) \: be \: a \: differntiable \: function \: and \: increasing$

Also,

$\rm :\longmapsto\:g(x) = {(f(x))}^{3} - 3 {(f(x))}^{2} + 4f(x) + 3sinx + 4cosx + 5x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}g(x) =\dfrac{d}{dx}\bigg[ {(f(x))}^{3} - 3 {(f(x))}^{2} + 4f(x) + 3sinx + 4cosx + 5x\bigg]$

$\rm :\longmapsto\:g'(x) = 3 {(f(x))}^{2}f'(x) - 6f(x)f'(x) + 4f'(x) + 3cosx - 4sinx + 5$

Now, as f(x) is increasing function.

$\rm \implies\:f'(x) \geqslant 0 - - - (1)$

Consider,

$\rm :\longmapsto\:3 {(f(x))}^{2}f'(x) - 6f'(x)f(x) + 4f'(x)$

$\rm \: = \:3f'(x)\bigg[ {(f(x))}^{2} - 2f(x)\bigg] + 4f'(x)$

$\rm \: = \:3f'(x)\bigg[ {(f(x))}^{2} - 2f(x) + 1 - 1\bigg] + 4f'(x)$

$\rm \: = \:3f'(x)\bigg[ {(f(x) - 1)}^{2} - 1\bigg] + 4f'(x)$

$\rm \: = \:3f'(x) {[f'(x) - 1]}^{2} - 3f'(x) + 4f'(x)$

$\rm \: = \:3f'(x) {[f'(x) - 1]}^{2} + f'(x)$

Hence,

$\bf\implies \:\:3 {(f(x))}^{2}f'(x) - 6f'(x)f(x) + 4f'(x) \geqslant 0$

Now,

We know that,

$\red{\rm :\longmapsto\:asinx + bcosx + c \in \: [c - \sqrt{ {a}^{2} + {b}^{2}},c + \sqrt{ {a}^{2} +{b}^{2}}]}$

Thus,

$\red{\rm :\longmapsto\:3sinx - 4cosx + 5 \in \: [5 - \sqrt{ {3}^{2} + {( - 4)}^{2}},5 + \sqrt{ {3}^{2} +{( - 4)}^{2}}]}$

$\red{\rm :\longmapsto\:3sinx - 4cosx + 5 \in \: [5 - \sqrt{ 9 + 16},5 + \sqrt{ 9 + 16}]}$

$\red{\rm :\longmapsto\:3sinx - 4cosx + 5 \in \: [5 - \sqrt{25},5 + \sqrt{25}]}$

$\red{\rm :\longmapsto\:3sinx - 4cosx + 5 \in \: [5 - 5, \: 5 + 5]}$

$\red{\rm :\longmapsto\:3sinx - 4cosx + 5 \in \: [0, \: 10]}$

$\red{\bf :\longmapsto\:3sinx - 4cosx + 5 \geqslant 0}$

So, from above calculations we have

$\bf :\longmapsto\: \:\:3 {(f(x))}^{2}f'(x) - 6f'(x)f(x) + 4f'(x) \geqslant 0$

and

$\red{\bf :\longmapsto\:3sinx - 4cosx + 5 \geqslant 0}$

So,

$\rm :\longmapsto\:g'(x) = 3 {(f(x))}^{2}f'(x) - 6f(x)f'(x) + 4f'(x) + 3cosx - 4sinx + 5$

$\bf\implies \:g'(x) \geqslant 0$

$\bf\implies \:g \: is \: increasing \: whenever \: f \: is \: increasing.$

Additional Information :-

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