Question

Calculus help?
analyze and sketch a graph of the function.
Label any intercepts, relative extrema, points of inflection, and
asymptotes

y= (-1/3) (x^3-3x+2)

Answer 1

can you find the first derivative ?

Answer 2

[phi] has the path

Answer 3

yeah. it is: -(3x^2-3)/3
sorry I wasn't on

Answer 4

you can simplify that to
f'(x) = -x^2+1 or 1-x^2

it might be useful to find the 2nd derivative f''= -2x

next, find the critical points, which is where f'(x)=0
in other words solve
1-x^2 =0

Answer 5

x= 1 would be the critical points right?

Answer 6

you solve x^2= 1
x= ±\(\sqrt{1}\)= ±1

so you get 2 critical points

Next make a table
x f(x) f'(x) f''(x)


solve for f(x) (that is, the y value) at the critical points
f' will be zero at those pts. evaluate f'' at the critical pts. a - # for f'' means a local max, a positive f'' means a local min.

put in some x values near the critical pts, say -2, 0, and 2
find f(x) at this points

plot the points

sketch in the curve between the pts.

next, notice what f(x) is as x -> very negative (it zooms up)
and what f(x) does as x-> very positive (it zooms down)

finish the sketch by showing this "end behavior"

Answer 7

make a table with the critical values of each function?

Answer 8

oh no never mind I understand now. got it thank you so much!