Extremum points?

Question

Find the extremum points, and specify if they are maximum, minimum points or no conclusions. 

1. Y= (2x)^2 

2. Y= (x^3) - ((2x)^2) - 3 

3. Y = ln((3x)^2) + 2)

Answer 1

(1)

 Y= (2x)² 

 Y = 4 x²

To find the maximum / minimum points equate y' = 0 and find for x.

Y' = 8x                   [ derivative of  xⁿ = n x^n-1 ]

8x = 0

x = 0

At  x = 0 the function y has maximum / minimum point 

To find the whether the function has maximum / minimum at x = 0 , find y'' and substitute the critical point ( x=0 ) .

Now find y'' 

y'' = 8 

Since y'' > 0 , the function y has minimum value at x = 0 .

Now put x = 0 in Y = 4 x²

y = 4 *0²

y = 0

So the minimum point is (0,0) .

Answer 2

(2)

 Y= x³ -(2x)² -3

 Y = x³ - 4x² -3

To find the maximum / minimum points equate y' = 0 and find for x.

Y' = 3x² -8x                   [ derivative of  xⁿ = n x^n-1 ]

3x² -8x = 0

x(3x - 8) = 0

x = 0 , x = 8/3

So the critical points are x = 0 , x = 8/3

At  x = 0 the function y has maximum / minimum point 

To find the whether the function has maximum / minimum at x = 0 , find y'' and substitute the critical point ( x=0 ) .

Now find y'' 

y'' = 6x 

y'' = 6(0) = 0

Since y'' = 0 , the function y has maximum value at x = 0 .

Now put x = 0 in Y = x³ - 4x² -3

y = 0³ - 4 *0² -3

y = -3

So the maximum point is (0,-3) .

At  x =  8/3  the function y has maximum / minimum point 

To find the whether the function has maximum / minimum at x =  8/3 , find y'' and substitute the critical point ( x= 8/3) .

Now find y'' 

y'' = 6x 

y'' = 6( 8/3) = 16

Since y'' > 0 , the function y has minimum value at x =  8/3.

Now put x =  8/3  in Y = x³ - 4x² -3

y = ( 8/3)³ - 4 *( 8/3)² -3

y = -12.48

So the manimum point is (0,-12.5) .

At (0,-3)  function y has maximum and at (0,-12.5) the function y has minimum  .

Answer 3

(3)

 Y= ln((3x)² + 2) 

 Y =ln( 9x² + 2 ) 

To find the maximum / minimum points equate y' = 0 and find for x.

Y' = 1/( 9x² + 2 )( 9x² + 2 )'                   [ derivative of  ln x = 1/x ]

Y' = 1/( 9x² + 2 )( 18x )

Y' = 18x/( 9x² + 2 )

18x/( 9x² + 2 ) = 0

18x = 0

x = 0

At  x = 0 the function y has maximum / minimum point 

To find the whether the function has maximum / minimum at x = 0 , find y'' and substitute the critical point ( x=0 ) .

Now find y''

Y' = 18x/( 9x² + 2 ) 

The quotient rule of derivative is    d [ u/v ] = (vu' - uv') / v² 

y'' = [ ( 9x² + 2 )( 18 ) - 18x ( 18x ) ]/( 9x² + 2 )²

Now put x = 0

y'' = [ ( 9(0)² + 2 )( 18 ) - 18(0) ( 18(0) ) ]/( 9(0)² + 2 )²

y'' = [(36 -0 )/4]

y'' = 9

Since y'' > 0 , the function y has minimum value at x = 0 .

Now put x = 0 in Y = ln( 6x² + 2 )

y = ln( 6(0)² + 2 )

y = ln (0+2)

y = ln 2

y = 0.693

So the minimum point is (0,0.693) .