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2 X X 2 X X X 2 X X 1 2 X X 2 X X X 2 X

The document contains information about two functions: 1) f(x)=x5-8x4+16x3+10. It finds the critical points, calculates the second derivative test at those points, and evaluates the original function at certain values. 2) f(x)=x2ex. It finds the derivative and second derivative, sets the second derivative equal to 0 to find critical points, and evaluates the original function at those points.

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Daniel Santander
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58 views2 pages

2 X X 2 X X X 2 X X 1 2 X X 2 X X X 2 X

The document contains information about two functions: 1) f(x)=x5-8x4+16x3+10. It finds the critical points, calculates the second derivative test at those points, and evaluates the original function at certain values. 2) f(x)=x2ex. It finds the derivative and second derivative, sets the second derivative equal to 0 to find critical points, and evaluates the original function at those points.

Uploaded by

Daniel Santander
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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f(x)=x5-8x4+16x3+10 x2(5x2-32x+48)

f´(x)=5x4-32x3+48x2 x1=0

f´´(x)=20x3-96x2+96x x2= 4 x3= 2.4

f´´(0)=20x3-96x2+96x= 0 f´´(4)=20x3-96x2+96x= 128 f´´(2.4)=20x3-96x2+96x= -46.08

f(4)=x5-8x4+16x3+10= -10 f(2.4)=x5-8x4+16x3+10= 25.38 f(0)=x5-8x4+16x3+10= -10 (4,-10)


(2.4,25.38) (0,-1)

f(x)=x2ex f´(x)=2xex+x2ex f´´(x)=2ex+4xex+x2ex xex(2+x) x1=0 x2= -2


f´´(-2)=2ex+4xex+x2ex = -0.2706 maximo f´´(0)=2ex+4xex+x2ex = 2 minimo
f(-2)=x2ex =0.54 f(0)=x2ex = 0 (-2,0.54) (0,0)

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