Question

(6).
Examine the function f(x)= x3 - 5x2 + 8x - 4 for maxima and minima.​

Answer 1

To examine:

The function f(x)= x3 - 5x2 + 8x - 4 for maxima and minima.

Calculation:

$f(x) = {x}^{3} - 5 {x}^{2} + 8x - 4$

$\implies \: {f}^{'} (x) = \dfrac{d \: f(x)}{dx}$

$\implies \: {f}^{'} (x) = \dfrac{d( {x}^{3} - 5{x}^{2} + 8x - 4)}{dx}$

$\implies \: {f}^{'} (x) = 3 {x}^{2} - 10x + 8$

For Maxima/Minima , f'(x) = 0

$\implies \: 3 {x}^{2} - 10x + 8 = 0$

$\implies \: 3 {x}^{2} - 6x - 4x + 8 = 0$

$\implies \: 3x(x - 2) - 4( x - 2) = 0$

$\implies \: (3x - 4)( x - 2) = 0$

Either x = 4/3 or 2.

Now, calculating f"(x):

$\implies \: {f}^{'} (x) = 3 {x}^{2} - 10x + 8$

$\implies \: {f}^{"} (x) = 6x - 10$

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$\implies \: {f}^{"} (x) \bigg|_{x = \frac{4}{3} } = 6( \frac{4}{3} ) - 10$

$\implies \: {f}^{"} (x) \bigg|_{x = \frac{4}{3} } = - 2$

$\implies \: {f}^{"} (x) \bigg|_{x = \frac{4}{3} } < 0$

Hence maxima at x = 4/3.

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$\implies \: {f}^{"} (x) \bigg|_{x = 2 } = 6( 2) - 10$

$\implies \: {f}^{"} (x) \bigg|_{x = 2 } = 2$

$\implies \: {f}^{"} (x) \bigg|_{x = 2 } > 0$

Hence , minima at x = 2.

Answer 2

Answer:

Step-by-step explanation: