Simple Calculus questions

Question

1) y = 2x^3 - 3x^2 has a local max value at x =?

2) Consider the sationary point (0,1) on y = x^3 - 3x^2 + 1

This is a: Local max, local min or neither

 

Answer 1

(1)

The function is f(x) = 2x³ -3x²

f '(x) = 6x² - 6x

f ''(x) = 12x - 6

 

Find Extrema :

To find out extrema, use theorem.

If f " (x) > 0 (positive) ------> minimum point.

If f " (x) < 0 (negative) ------> maximum point.

f '(x) = 6x² - 6x = 6x(x - 1) = 0.

The roots are x = 0 and x = 1

So, lets plug each critical point in f " (x) = 12x - 6.

If x = 1 then f " (1) = 12(1) - 6 = 12 - 6 = 6 > 0 (positive), therefore local minimum point.
If x = 0 then f " (0) = 12(0) - 6 = 0 - 6 = -6 < 0 (negative), therefore local maximum point.

To find the f(x) to each x for local max and local min plugging those values in the original function.

If x = 0 then, f(0) = 2(0)³ - 3(0)²  = 0.

The relative maximum point is (0, 0). 

Answer 2

(2)

The function is f(x) = x³ -3x²+1

f '(x) = 3x² - 6x

f ''(x) = 6x - 6

 

Find Extrema :

To find out extrema, use theorem.

If f " (x) > 0 (positive) ------> minimum point.

If f " (x) < 0 (negative) ------> maximum point.

f '(x) = 3x² - 6x = 0

3x(x-2) = 0

x = 0 or x = 2

The roots are x = 0 and x = 2

If x = 0 then f " (0) = 6(0) - 6 = -6< 0 (negative), therefore local maximum point.

To find the f(x) to each x for local max and local min plugging those values in the original function.

If x = 0 then f(x) = (0)³ -3(0)² +1  = 1

Therefore the point (0,1) is local maximum point