Find relative extrema of f(x)=x^3-3x^2+4

Answer 1

To find relative maxima and minima, first find the critical points (where f�Œ is 0 or doesn�ft exist).

Then examine each critical point.

It is a relative maximum if f�Œ changes from positive to negative or f�� is negative.

It is a relative minimum if f�Œ changes from negative to positive or f�� is positive.

We find the critical numbers of f by solving the equation f'( x ) = 0

f'( x ) = 3 * x ^ 2 - 2 * 3 * x

f'( x ) = 3 x ^ 2 - 6 x

f'( x ) = 3 x ( x - 2 ) = 0

So the critical numbers are x = 0 and x = 2

f�� ( x ) = 3 * 2 * x - 6 = 6 x - 6 = 6 ( x - 1 )

For x = 0

f�� = 6 * ( 0 - 1 ) = 6 * - 1 = - 6

f�� < 0

It is a relative maximum

For x = 2

f�� = 6 * ( 2 - 1 ) = 6 * 1 = 6

It is a relative minimum

Relative maximum :

x = 0

y = 0 ^ 3 - 3 * 0 ^ 2 + 4 = 4

Relative minimum :

x = 2

y = 2 ^ 2 - 3 * 2 ^ 2 + 4 = 8 - 3 * 4 + 4 = 8 - 12 + 4 = 0

If you want to see graph In google type:

functions graphs online

When you see list of results click on:

rechneronline.de/function-graphs/

When page be open in blue rectangle type:

x^3-3x^2+4

Then click option Draw

Answer 2

To find relative maxima and minima, first find the critical points (where f´ ( x ) is 0 or doesn´t exist).

Then examine each critical point.

It is a relative maximum if f´ changes from positive to negative or f" is negative.

It is a relative minimum if f´ changes from negative to positive or f" is positive.

f�� mean f "