Question

The greatest value of f(x)=2x³ - 2x² - 12x+ 1 in the interval [-2, 5] is​

Answer 1

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

Given function is

$\rm \: f(x) = {2x}^{3} - {2x}^{2} - 12x + 1$

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

$\rm \: \dfrac{d}{dx}f(x) = \dfrac{d}{dx}[ {2x}^{3} - {2x}^{2} - 12x + 1 ]$

We know,

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

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

So, using these results, we get

$\rm \: f'(x) = {6x}^{2} - 4x - 12$

For critical points,

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

$\rm \: {6x}^{2} - 4 x - 12 = 0$

On dividing by 2 both sides,

$\rm \: {3x}^{2} - 2 x - 6 = 0$

Now, its a quadratic equation, so we use Quadratic Formula to get the values of x

So, by using Quadratic formula, we get

$\rm \: x = \dfrac{ - ( - 2) \: \pm \: \sqrt{ {( - 2)}^{2} - 4(3)( - 6)} }{2(3)}$

$\rm \: x = \dfrac{ 2 \: \pm \: \sqrt{ 4 + 72} }{6}$

$\rm \: x = \dfrac{ 2 \: \pm \: \sqrt{76} }{6}$

$\rm \: x = \dfrac{ 2 \: \pm \: 2\sqrt{19} }{6}$

$\rm \: x = \dfrac{ 1 \: \pm \: \sqrt{19} }{3}$

$\rm \: x = \dfrac{ 1 \: \pm \: 4.36 }{3}$

$\rm \: x = \dfrac{ 5.36 }{3} \: \: or \: \: x = \dfrac{ - 3.36}{3}$

$\rm\implies \:x = 1.79 \: \: or \: \: x = - 1.12 \:\:\:\in \:[-2, 5]$

So, stationary points are x = - 2, - 1.12, 1.79, 5

Now, let calculate the value of f(x) at stationary points

We have,

$\rm \: f(x) = {2x}^{3} - {2x}^{2} - 12x + 1$

So,

$\rm \: f( - 2) = {2( - 2)}^{3} - {2( - 2)}^{2} - 12( - 2) + 1$

$\rm \: f( - 2) = - 16 - - 8 + 24 + 1$

$\rm\implies \:f( - 2) = 1$

Now,

$\rm \: f( - 1.12) = {2( - 1.12)}^{3} - {2( - 1.12)}^{2} - 12( - 1.12) + 1$

$\rm \: f( - 1.12) = - 2.809856 - 2.5088 + 13.44 + 1$

$\rm\implies \:f( - 1.12) = 9.12 \: ( \: approx \: )$

Now,

$\rm \: f(1.79) = {2(1.79)}^{3} - {2(1.79)}^{2} - 12(1.79) + 1$

$\rm \: f(1.79) = 11.4711 - 6.408 - 21.48 + 1$

$\rm\implies \:f(1.79) = - \: 15.42 \: \: ( \: approx \: )$

Now,

$\rm \: f(5) = {2(5)}^{3} - {2(5)}^{2} - 12(5) + 1$

$\rm \: f(5) = 250 - 50 - 60 + 1$

$\rm\implies \:f(5) = 141$

So, from the above calculations, we concluded that

Greatest value of f(x)=2x³ - 2x² - 12x+ 1 in the interval [-2, 5] is 141 at x = 5

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Additional Information :-

Method to find Local maxima or Local minima  :-

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.

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