Question

Find the maximum and minimum value of the function and function
f(x)=x³-2x²+x+9

Answer 1

Step-by-step explanation:

Given:

$f(x) = {x}^{3} - 2 {x}^{2} + x + 9$

To find: Find the maximum and minimum value of the function.

Solution:

Tip:Apply second derivative test

Step 1: Find f'(x)

$f'(x) = 3 {x}^{2} - 4x + 1$

Step 2: Put f'(x)=0 and find value of x

$3 {x}^{2} - 4x + 1 = 0 \\ \\ 3 {x}^{2} - 3x - x + 1 = 0 \\ \\ 3x(x - 1) - 1(x - 1) = 0 \\ \\ (x - 1)(3x - 1) = 0 \\ \\ x = 1 \\ \\ or \\ \\ x = \frac{1}{3}$

Step 3: Find f"(x)

$f"(x) = 6x - 4$

Step 4: Put x=1 and x=1/3 in f"(x)

$f''(1) = 6(1) - 4 \\ \\ f''(1) = 2$

f"(1)>0;implies there is minima at x=1.

$f''( \frac{1}{3} ) = 6 \times \frac{1}{3} - 4 \\ \\ f''( \frac{1}{3} ) = - 2$

f"(1/3)<0; there is maxima at x=1/3.

Step 5: Maximum and minimum values of function

$f(1) = {(1)}^{3} - 2 {(1)}^{2} + 1 + 9$

$f(1) = 1 - 2 + 1 + 9$

$f(1) = 9$

$f( \frac{1}{3} ) = {( \frac{1}{3} )}^{3} - 2 {( \frac{1}{3} )}^{2} + \frac{1}{3} + 9$

$f( \frac{1}{3} ) = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} + 9$

$f( \frac{1}{3} ) = \frac{1 - 6 + 9 + 243 }{27}$

$f( \frac{1}{3} ) = \frac{247}{27}$

$f( \frac{1}{3} ) =9.15$

Final answer:

Minima is at x=1 and minimum value is 9.

Maxima is at x=1/3 and maximum value is 9.15

Hope it helps you.

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