Help please !! f(x)=−2x^3−6x^2+210x−6 is increasing on the interval ( ....... , ....... ) ?

Question

f(x)=−2x^3−6x^2+210x−6

is increasing on the interval ( ....... , ....... ) ?

It is decreasing on the interval ( −∞, .......  ) and the interval (  ...... , ∞ ). ?

The function has a local maximum at  ....... ?

Answer 1

The first step in solving this problem is to find the critical points. The critical points are our candidates for local minima and maxima. This is done by setting the first derivative equal to zero and solving.

Thus, for the first derivative, we have: f'(x) = -6x²-12x+210.

Setting this equal to zero and solving:

-6x²-12x+210=0

x+2x-35=0

(x+7)(x-5)=0

x=-7,5

Our critical points are then at x equals -7 and 5 and plugging them into f(x) gives us (-7,-1084) and (5,644). At the critical points, f'(x) is equal to zero, meaning f(x) is neither increasing nor decreasing. At all other points, f'(x) is not equal to zero and f(x) must be increasing/decreasing until it reaches a critical point where it then switches behavior.

To determine the intervals of increase/decrease, we find the value of f'(x) around -7 and 5. Accordingly, let's test the points x = -8, 0, 6 which gives:

f'(-8) = -78

f'(0) = 210

f'(6) = -78

This means that on the intervals (-∞,-7) and (5,∞), f(x) is decreasing as f'(x) is negative in these regions and on the interval (-7,5), f(x) is increasing as f'(x) is positive.

For a critical point to be a local maximum, f(x) must be increasing towards the critical point and then decreasing after it passes the critical point. We see this for the point (5,644) as f(x) is increasing in the interval (-7,5) and then begins to decrease once it passes x=5 in the interval (5,∞). Thus, (5,644) is a local maxima.

(For there to be a local minimum, the function must decrease towards the critical point and then increase once it passes it. Accordingly, (-7,-1084) is a local minimum.)