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^2x2/x^2+x6
i did this and got my answers but i doubt they are right, can anyone get the answers if you have a TI8384 please? thank you.

precalculus  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 xcoordinate of the maximum/minimum to see if f'(x)=0. This will be another way to confirm your answer.

precalculus  Paris, Sunday, March 14, 2010 at 5:46pm
my answer is :
maxima: x= 1.56
minima:x=1.29, x=1.25

precalculus  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*x12/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.

precalculus  Paris, Sunday, March 14, 2010 at 11:40pm
oh thank you so much, so x=1.56 is a maxima, and we don't have a minima?
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