Posted by **Paris** on Saturday, March 13, 2010 at 9:46pm.

Also with calculator, graph f(x), and determine all the possible maxima/minima coordinates to within two decimal points.

f(x)=x^3+2x^2-x-2/x^2+x-6

i did this and got my answers but i doubt they are right, can anyone get the answers if you have a TI-83-84 please? thank you.

- pre-calculus -
**MathMate**, Saturday, March 13, 2010 at 10:21pm
Could you post your answers if you need a confirmation?

Also, calculate f'(x) and substitute the x-coordinate of the maximum/minimum to see if f'(x)=0. This will be another way to confirm your answer.

- pre-calculus -
**Paris**, Sunday, March 14, 2010 at 5:46pm
my answer is :

maxima: x= -1.56

minima:x=-1.29, x=-1.25

- pre-calculus -
**MathMate**, Sunday, March 14, 2010 at 8:51pm
f'(x)=3*x^2+4*x+4/x^3

f'(-1.56)=0.0072

f'(-1.559)=-0.00021

Therefore x=-1.56(approx.) is a maximum or minimum.

f"(x)=6*x-12/x^4+4

f"(-1.56)=-7.386 >0

Therefore x=-1.56 is a maximum.

f'(-1.29)=-2.03 ≠ 0

f'(-1.25)=-2.3605 ≠ 0

Therefore x=-1.29 and x=-1.25 are not maxima nor minima.

Check your graph without forgetting that the function is undefined at x=0.

- pre-calculus -
**Paris**, Sunday, March 14, 2010 at 11:40pm
oh thank you so much, so x=-1.56 is a maxima, and we dont have a minima?

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