Mathamatics

1 )Find the absolute maximum and minimum values of

f (x x3-3x 1 on the interval x is either less than or equal to 3 or greater than or equal to 0

Solution

Get the derivative

f (x 3 x2-3

Equate to 0 and solve for x

0 3 x2-3

x2 3 /3

The square root of 1 is 1 and -1 , therefore the two values are

x1 1

x2 -1

This is confirmed by the graph shown below

Solve for the intervals that is less than [banner_entry_middle]

or equal to 3 or greater than or equal to 0

Maximum

fmax f (-1 (-1 )3-3 (-1 1 3

Minimum

fmin f (1 (1 )3-3 (1 1 -1

2 ) find the local maximum and minimum values of

f (x x5 – 5x 3 using

a ) The first derivative test

Solution

Get the derivative

f (x 5×4 – 5

Equate to 0 and solve for x

0 5×4 – 5

x4 (5 /5

x 1 or -1

1st Derivative test for x 1

f (0 .9 -1 .72

f (1 .1 2 .32

Since f (0 .9 ) is negative and f (1 .1 ) is positive then it is a local minimum

1st Derivative test for x -1

f (-1 .1 2 .32

f (-0 .9 -1 .72

Since f (-1 .1 ) is positive and f (-0 .9 ) is negative then it is a local maximum

b ) The second derivative test

Solution

Get the second derivative

f (x 20x

2nd Derivative test for x 1

f (1 20

Since f (x 0 then it is local minimum

2nd Derivative test for x 1

f (-1 -20

Since f (x 0 then it is local maximum

c ) Which do you Prefer ? Why

I like the second derivative test since it is easier to evaluate and compare

3 ) Find the most general antiderivative (add a constant ) of f (x 3×2 1 / x (where x is greater than 0

Solution

Ax (f (x F (x (3 /3 )x3 lnx C

F (x x3 lnx C

4 (a ) Find the most general antiderivative F of f (x 1 – sin x

Solution

F (x x – (-cosx C

F (x x cosx C (b ) Find the Antiderivative which satisfies

F (0 0

Solution

0 0 cos (0 C

C – cos (0

C -1

Therefore the equation is

F (x x cosx – 1… [banner_entry_footer]