please check work on 2x^3 - 3x^2 - 12x + 5

[tkthustler]

2x^3 - 3x^2 - 12x + 5

f' < 0 on interval (-1, 2)
f' >0 on intervals (-infinity, -1) and (2, infinity)

f' = 0 at x = 2 and at x = -1

f is concave upward on the interval (1/2, infinity)
f is concave downward on the interval (-infinity, 1/2)

local maxima at (-1, 2)
inflection point at x = -1/2

Also, what does it mean to graph f?

Any help is appreciative. Thank you so much!

 

[stapel]

What were the instructions? Which parts of your post is problem-statement and which is your working?

"Graphing" just means that: pick x-values, plug them into f, find the corresponding y-values, draw the dots, and (together with what you have on intercepts, max/min points, concavity, and inflection points) sketch in the line for f(x).

Eliz.

 

[galactus]

You mean you don't have a graphing calculator?. You've never ever graphed anything?. A graph is a great way to check yourself with these types of problems.

Now, you can easily see the concaves.