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]
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