Question

the maximum and minimum value of the following function
$y = {4x}^{3} - {18x}^{2} + 24x - 7$
​

Answer 1

Answer:

14x-19y^5 ok

Step-by-step explanation:

bro understand

Answer 2

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

Given function is

$\bf :\longmapsto\:y = {4x}^{3} - {18x}^{2} + 24x - 7 - - - (1)$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx}( {4x}^{3} - {18x}^{2} + 24x - 7)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{4x}^{3} - \dfrac{d}{dx}{18x}^{2} +\dfrac{d}{dx} 24x - \dfrac{d}{dx}7$

We know that

$\red{\boxed{ \sf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

and

$\red{\boxed{ \sf \: \dfrac{d}{dx}k = 0}}$

So, using these results we get

$\bf :\longmapsto\:\dfrac{dy}{dx} = {12x}^{2} - 36x + 24 - - - (2)$

For maxima or minima,

$\rm :\longmapsto\:\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\: {12x}^{2} - 36x + 24 = 0$

$\rm :\longmapsto\: 12({x}^{2} - 3x + 2) = 0$

$\rm :\longmapsto\: {x}^{2} - 3x + 2= 0$

$\rm :\longmapsto\: {x}^{2} - 2x - x + 2= 0$

$\rm :\longmapsto\:x(x - 2) - 1(x - 2) = 0$

$\rm :\longmapsto\:(x - 2)(x - 1) = 0$

$\bf\implies \:x = 2 \: \: \: or \: \: \: x = 1$

Now, from equation (2),

$\rm :\longmapsto\:\dfrac{dy}{dx} = {12x}^{2} - 36x + 24$

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

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{d}{dx}( {12x}^{2} - 36x + 24 )$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{d}{dx}{12x}^{2} - \dfrac{d}{dx}36x + \dfrac{d}{dx}24$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = 24x - 36$

Now, Consider x = 1

$\rm :\longmapsto\:\bigg[\dfrac{ {d}^{2} y}{d {x}^{2} }\bigg]_{x = 1} = 24 - 36 = - 12 < 0$

$\bf\implies \:y \: is \: maximum \: at \: x = 1$

and

$\rm :\longmapsto\:Maximum \: value = 4 - 18 + 24 - 7 = 3$

Now Consider x = 2

$\rm :\longmapsto\:\bigg[\dfrac{ {d}^{2} y}{d {x}^{2} }\bigg]_{x = 2} = 48 - 36 = 12 > 0$

$\bf\implies \:y \: is \: minimum \: at \: x = 2$

and

$\rm :\longmapsto\:Manimum \: value = 4 {(2)}^{3} - 18 {(2)}^{2} + 24(2) - 7$

$\rm :\longmapsto\:Manimum \: value = 32 - 72 + 48 - 7 = 1$

$\red{\rm :\longmapsto\:\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{Maximum \: value = 3} \\ &\sf{Minimum \: value = 1} \end{cases}\end{gathered}\end{gathered}}$

Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

The function f (x) is maximum when f''(x) < 0.

The function f (x) is minimum when f''(x) > 0.

Additional Information :-

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