# pre-calculus

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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 - ,

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 - ,

my answer is :

maxima: x= -1.56
minima:x=-1.29, x=-1.25

• pre-calculus - ,

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 - ,

oh thank you so much, so x=-1.56 is a maxima, and we don't have a minima?

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