Question

Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.)

(a). f(θ) = 12 cos θ + 6 sin²θ

(b). h(p) = p-2/p2+9

(c). f(x) = x⁵e^−6x

^(d). f(x) = x⁻⁹ ln x

(e). If f(3) = 3 and f '(x) ≥ 1 for 3 ≤ x ≤ 5, how small can f(5) possibly be?

Answer 1

a) Consider the function, f(9)-12 cos θ + 6 sin, θ Differentiate/with respect to θ -12 sin θ + 12sin θ cos θ -12sin θ(-1 + cos θ) To find the critical numbers, solve f,(9)-0 12sino(-1 + cos θ)-0 sin θ or cos θ-1 Hence, the critical numbers ofthe functionfare8-2ηπ integer or θ-π+2ηπ, where n is an b) Consider the function, Differentiate h with respect to p -2)-(P2-9) (p+9) (p+9 1)-(p-2) (2p) (p+9) (p +9)

To find the critical numbers, solve f(x)- 0 x"e-"(-6x + 5)=0 x0 or 6x+5-0 x"=0 or 6x=5 Hence, the critical numbers of the function,fare x=0 and x-

e) Consider that f(3)3 and f (x)>1 for 3Sxs5 Since f(x) 1 for 3sx S5, note that the function is always have positive slope Therefore, the function fis increasing in the interval 33xs5. The minimum slope of the function is 1 So now consider the minimum slope f(x)-1 Integrate on both sides f(x)-x+C Use the initial condition f (3) 3 Substitute x 3 andf 3 in f (x)-x+C f (3)-3+C 3 3+C С 0 Therefore, the function is f(x)=x substitute x=5 in f(x) Thus, f (5) 5 is the possible small value.[