Let me know if it\'s to small to understand. Consider the function f(x) = x4 -6x

Question

Let me know if it's to small to understand.

Consider the function f(x) = x4 -6x2 +8x + 10 What is the domain of the function? Express as an interval. Find the derivative of the function. Find x where f'(x) = 0 or f'(x) does not exist. Determine at each of the points in part {c}, whether the function has a local minimum or maximum Find any absolute maximum and/or minimum values of the function and where it occurs. Consider the function What is the domain of the function? Express as an interval. Show that the derivative Find x where f'(x) = 0 or f'(x) does not exist. Determine at each of the points in part {c}. whether the function has a local minimum or maximum.

Explanation / Answer

4) f(x)=x^4-6x^2+8x+10

a) Domain is R i.e (-infinity,infinity) as it is a polynomial

b)f'(x)=4x^3-12x+8=4(x^3-3x+2)=4(x-1)(x^2+x-2)=4(x-1)(x+2)(x-1)

c)So f'(x)=0 at x=1 or -2

d)By first derivative test we see there is no sign change about x=1. so no local max/min there

at x=-2 sign changes from - to + So local min at x=-2

e) f(-2)=-14 ,f(infinity)=f(-infinity)=infinity

So absolute max at x=infinity

and absolute min at x=-2

5)f(x)=x^2*sqrt(x+2)

a)domain is when x>=-2 i.e [2,infinity) such that term in sqrt root>=0

b)f'(x)=2xsqrt(x+2)+x^2/(2sqrt(x+2))=(5x^2+8x)/(2sqrt(x+2))=x(5x+8)/(2sqrt(x+2))

c)f'(x) does not exist at x=-2

f'(x)=0 at x=0 and x=-8/5=-1.6

d)By first derivative test we see sign change from - to + about x=0. so local min there

and about x=-1.6 sign changes from + to - So local max at x=-1.6

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