Question

Find the maximum and minimum values of function: x^3-2x^2+x+6

Answer 1

Answer:

$let \: y = x^{3} - 2x^{2} + x + 6 \\ \frac{dy}{dx} = 3x^{2} - 4x + 1 \: and \: \frac{d ^{2} y}{d ^{2}x } = 6x - 4 \\ max \: nd \: min \: value \: of \: y \: nd \: \frac{dy}{dx} = 0 \\ 3 {x}^{2} - 4x + 1 = 0 \\ 3 {x}^{2} - 3x - x + 1 = 0 \\ 3x(x - 1) - 1(x - 1) = 0 \\ (3x - 1)(x - 1) \\ x = 1 \: and \: \frac{1}{3} \\ again \: \frac{d ^{2} y}{d^{2}x } at \: x = 1 \\ \frac{d ^{2}y }{d ^{2}x } = 6x - 4 \\ 6 \times 1 - 4 = 2 > 0 \\ x = \frac{1}{3} \\ maximum \: and \:minimum \: value \: at \: x = \frac{1}{3} \: and \: y = 1$

Hope it will help you

Answer 2

Answer:

maximum is 6.15 and minimum is 6

Step-by-step explanation:

given:

function f(X) = x^3-2x^2+x+6

to know maximum and minumum values, we need to find for which value of X, f(X) is at extremes.

for that we need to differentiate it and equate it to 0.

d(f(X))/dx = 3x²-4x+1

3x²-4x+1=0

3x²-3x-x+1=0

3x(x-1)-1(x-1)=0

(3x-1)(x-1) =0

X= 1/3 or x = 1

now substitute these values in f(X)

f(1/3) = (1/3)³ -2(1/3)²+(1/3)+6

= 1/27 - 2/9+1/3+6

= (6*27 +9-(3*2)+1)/27

= (162+9-6+1)/27

= 166/27

=6.15

f(1) = 1³-2(1)²+1+6

= 1-2+1+6

= 6