Question

Determine critical points of f(x) =8x³ + 81x²-42x-8

Answer 1

Step-by-step explanation:

Given:f(x) =8x³ + 81x²-42x-8

To find: Determine critical points of f(x)

Solution:

Step 1: Differentiate f(x)

$f'(x) = 24 {x}^{2} + 162x - 42$

Step 2: Put f'(x)=0,and find the values of x

$24 {x}^{2} + 162x - 42 = 0$

or

$12 {x}^{2} + 81x - 21 = 0$

apply quadratic formula

$x = \frac{ - 81 ± \sqrt{6561 + 1008} }{24} \\ \\ x = \frac{ - 81 ± \sqrt{ 7569} }{24} \\ \\ x = \frac{ - 81 ± 87}{24} \\ \\ x = \frac{1}{4} \\ \\ x = - 7$

Step 3: Find f"(x)

$48x + 162$

Step 4: Put values of x in f"(x)

$48 \times \frac{1}{4} + 162 \\ \\ 21 + 162 \\ \\ 183 > 0$

At x=1/4, minima is present.

Check for -7

$48( - 7) + 162 \\ - 336 + 162 \\ - 174 < 0$

at x=-7,there is a maxima.

Final answer:

Critical values of f(x) are x=1/4 and x= -7.

Hope it helps you.

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Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given:f(x) =8x³ + 81x²-42x-8

To find: Determine critical points of f(x)

Solution:

Step 1: Differentiate f(x)

$f'(x) = 24 {x}^{2} + 162x - 42$

Step 2: Put f'(x)=0,and find the values of x

$24 {x}^{2} + 162x - 42 = 0$

or

$12 {x}^{2} + 81x - 21 = 0$

apply quadratic formula

$x = \frac{ - 81 ± \sqrt{6561 + 1008} }{24} \\ \\ x = \frac{ - 81 ± \sqrt{ 7569} }{24} \\ \\ x = \frac{ - 81 ± 87}{24} \\ \\ x = \frac{1}{4} \\ \\ x = - 7$

Step 3: Find f"(x)

$48x + 162$

Step 4: Put values of x in f"(x)

$48 \times \frac{1}{4} + 162 \\ \\ 21 + 162 \\ \\ 183 > 0$

At x=1/4, minima is present.

Check for -7

$48( - 7) + 162 \\ - 336 + 162 \\ - 174 < 0$

at x=-7,there is a maxima.

Final answer:

Critical values of f(x) are x=1/4 and x= -7.

Hope it helps you.