Y= (1/2)x + cosx on the interval -2 pi ≤ x ≤ 2 pi. use the second derivative test ti find the local extrema of f

Step-by-step explanation:

y=(1/2)x+cos x

dy/dx=1/2-sin x

dy/dx=0,gives

1/2 -sin x=0

sin x=1/2=sin (π/6),sin (π-π/6)=sinπ/6,sin 5π/6=sin (2nπ+π/6),sin(2nπ+5π/6),n∈ I

d^2y/dx^2=-cos x>0 at x=π/6 and -2π+π/6

or at x=π/6,-11 π/6

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

the answer is c hope this