Question

For the function, f(x)=2x^3+3x^2-12x+1 use the first and second derivatives of f(x) to
find the following. (3 points each)
a. Find the intervals on which f is increasing or decreasing.
b. Find the local maximum and minimum values of f(x).
c. Find the intervals on which f is concave up and concave down.
d. Find any inflections points.
e. Find the maximum value of this function on [−2,4].

Answer 1

What did you find so far?

Answer 2

just started but

Answer 3

6x^2+6x-12

Answer 4

\(\large\color{black}{ \displaystyle f(x)=2x^3+3x^2-12x+1 }\)

\(\large\color{black}{ \displaystyle f'(x)=6x^2+6x-12 }\)

Answer 5

6(^2+x-2)

Answer 6

yes, I know what you meant (try to avoid such typos tho, please)

Answer 7

x=1, x=-2

Answer 8

Those are the roots of the derivative?

Answer 9

You need, though, to solve two cases;
\(\large\color{black}{ \displaystyle 0<6x^2+6x-12 }\)
(the solution to this will tell you where f increases (since the derivative at these values is positive))

\(\large\color{black}{ \displaystyle 0>6x^2+6x-12 }\)
(the solution to this will tell you where f decreases (since the derivative at these values is negative))

Answer 10

critical numbers

Answer 11

But you just need to find the intervals at which the function is increasing, and intervals at which the function is decreasing.

Answer 12

ok

Answer 13

that is increasing,
and decreasing would be:

0<6(x-1)(x+2)

x \(\in\) (-∞,-2) \(\cup\) (1,∞)


0>6(x-1)(x+2)

x \(\in\) (-2, 1)

Answer 14

Concave up, f''(x)>0
Concave down, f''(x)<0

Answer 15

Points of inflection
f''(x)=0

Answer 16

e)

- Find critical numbers (they are: -2,1,4)
- Find which one is larger, f(-2), f(1) or f(4) ?

Answer 17

i got everything exept c and d

Answer 18

i don't understand them