Consider the function f(x)=x√(x+8)

Question

a.) f(x) is decreasing on the interval

b.) f(x) is increasing on the interval

c.) f(x) has a local minimum at x=

d.) f(x) is concave up on the interval

Answer 1

a) f(x) is decreasing where f'(x) is negative

f'(x)=√(x+8)+x(1/(2√(x+8)) (Domain x>-8)

find zeros

0=(2(x+8)+x)/(2√(x+8)) Find common denominator

0=(3x+16)/(2√(x+8)) Simplify (combine like terms)6

0=3x+16 Set Numerator=0

x=-16/3

Do a sign test:

pick a number less than -16/3 such as -6

f'(-6)=(3(-6)+16)/(2√((-6)+8))

f'(-6)=(-2)/(2√(2) Numerator is negative, Denominator is positive. f'(-6) is negative

pick a number more than -16/3 such as 0

f'(0)=(3(0)+16)/(2√(0+8))

f'(0)=16/(2√8) Numerator and denominator are positive f'(0) is positive √

f(x) is decreasing from (-8,-16/3) (function does not exist less than -8)

b) f(x) is increasing where f'(x) is positive (-16/3,inf)

c) minimum is at x=-16/3

d) concave up where f''(x) is positive

f'(x)=(3x+16)/(2√(x+8))

f''(x)=(((2√(x+8))(3)-(3x+16)((x+8)^(-1/2)))/(4(x+8)

f''(x)=((6x+48)-(3x+16))/(√(x+8))(4(x+8))

set equal to zero

0=(3x+32)/(4√(x+8)^3/2)

x=-32/3≈-10.3 outside of domain

so concave up for entire domain since f''(x) is positive for entire domain (test value if not sure) f''(0)=+ and no zeros in domain of f''(x) so cannot be negative.