Question

Find the intervals of strictly increasing and decreasing for the function

$f(x) = {4sin}^{3}x - {6sin}^{2}x + 12sinx + 6 \: in \: (0 \: to \: \pi)$
​

Answer 1

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

Given function is

$\rm \: f(x) = {4sin}^{3}x - {6sin}^{2}x + 12sinx + 6$

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

$\rm \: \dfrac{d}{dx} f(x) = \dfrac{d}{dx}[ {4sin}^{3}x - {6sin}^{2}x + 12sinx + 6]$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }} \\ \\ \boxed{\tt{ \dfrac{d}{dx}sinx \: = \: cosx \: }}$

So, using these Identities, we get

$\rm \: f'(x) = 12 {sin}^{2}x(cosx) - 12sinx(cosx) + 12cosx + 0$

$\rm \: f'(x) = 12cosx[ {sin}^{2}x - sinx + 1 ]$

Now, we know that, for a quadratic expression f(x) = ax² + bx + c, if a > 0 and Discriminant, D = b² - 4ac < 0, then f(x) > 0.

Now, for the expression sin²x - sinx + 1, the coefficient of sin²x is 1 ( > 0 ) and Discriminant, D = 1 - 4 = - 3 < 0

So, it means

$\rm \: {sin}^{2}x - sinx + 1 > 0$

So, increasing and decreasing depends on sign of cosx.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \: of \: f'(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \bigg(0,\dfrac{\pi}{2}\bigg) & \sf + ve \\ \\ \sf \bigg(\dfrac{\pi}{2}, \: \pi\bigg) & \sf - ve \end{array}} \\ \end{gathered}$

So, For strictly increasing,

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

$\rm\implies \:x \: \in \: \bigg(0, \: \dfrac{\pi}{2}\bigg)$

And, For strictly decreasing,

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

$\rm\implies \:x \: \in \: \bigg(\dfrac{\pi}{2}, \: \pi\bigg)$

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ADDITIONAL INFORMATION

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

Answer 2

Given :

$\sf{f(x) = {4 \: sin}^{3}x - {6sin}^{2}x + 12sinx + 6 \: in \: (0 \: to \: \pi)}$

So,

$\sf{f'(x) = {12 \: sin}^{2}x.cos \: x + 12 \: cos \: x ]}$

$\sf{= 12 \: cos \: x[si {n}^{2} x - sin \: x + 1]}$

As -1 ≤ sin x ≤ 1,

$\sf\implies{ f'(x) = 12 \: cos \: x[si {n}^{2} x + (1 - sin \: x)]..(i) ]}$

As sin² x + 1-sinx ≥ 0

Therefore,

$\sf{1-sin \: x ≥ 0, also \: si {n}^{2} x ≥ 0}$

Hence ,

• f'(x)>0, when cos x>0 i.e.

$\sf\implies{ x \: ∈( \frac{ - \pi}{2}, \frac{\pi}{2}) }$

and

• f'(x) <0, when cosx <0 i.e.,

$\sf\implies{ x \: ∈( \frac{ \pi}{2}, \frac{3\pi}{2}) }$

When,

$\sf\implies{ x \: ∈( \frac{ - \pi}{2}, \frac{\pi}{2}) }$

• then f(x) is increasing

When,

$\sf\implies{ x \: ∈( \frac{ \pi}{2}, \frac{3\pi}{2}) }$

• then f(x) is decreasing

Since,

$\sf{( \frac{ - \pi}{2},\pi})∈( \frac{ \pi}{2}, \frac{3\pi}{2})$

• Therefore, f (x) is decreasing in

$\sf\implies{( \frac{ - \pi}{2},\pi})$