Find where the function is increasing and where it is decreasing. f (x) = (x + 2

Question

Find where the function is increasing and where it is decreasing. f (x) = (x + 2) (x - 6) Increasing on [-12, infinity), decreasing on (- infinity, -12] Decreasing on (- infinity, infinity) Increasing on [2, infinity), decreasing on (- infinity, 2] Increasing on (- infinity, -2] Union [6, infinity), decreasing on [-2, 6] h (t) = 1/t^2 + 1 Increasing on (- infinity, 0], decreasing on [0, infinity) Increasing on (- infinity, infinity) Increasing on [0, infinity), decreasing on (-infinity, 0] Increasing on (- infinity, 1], decreasing on [1, infinity) h (z) = 108z - z^3 Increasing on (- infinity, 6], decreasing on [6, infinity) Increasing on [- 36, 36], decreasing on (- infinity, - 36] Union [36, infinity) Increasing on [-6, 6], decreasing on (- infinity, -6] Union [6, infinity) Increasing on (- infinity, -6] Union [6, infinity), decreasing on [- 6, 6]

Explanation / Answer

12.

f(x)=(x+2)(x-6)

It is a quadratic equation and if we want it's root then

f(x)=(x+2)(x-6)=0

x=-2,6

and we look at the

f'(x)=x+2+x-6=2x-4=0

at x=2 we can have max or min so just check

f''(x)=2

so at x=2 we have global minimum

So decreasing on (-infinity, 2] and increasing on [2,infinity)

13.

h(t)=1/(t^2+1)

We can say that h(t) will be always positive for any value of t.

So at t=0, h(0)=1 and either t increases or decreases h(t) will decrease

as we approch t=-infinite to 0, value increase and as we approch t= 0 to infinite, value decrease.

increase on (-infinite, 0] and decrease in [0, infinite)

14

h(z)=108z-z^3=z(108-z^2)

h'(z)=108-3z^2=0; z=6, -6

h''(z)=-6z

at z=6 h''(Z)=-36, -ve so at z=6 we got maximum

at z=-6 h''(Z)=36, +ve so at z=-6 we got minimum

so it is increasing on [-6,6]

h(-7)=-413

h(-6)=-432

h(6)=432

h(7) =413

Option C is correct

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