which intervals is function f(x0=12x^2/3-4x increasing

Question

A) find the intervals on which f is increasing 

 

 

B) Find the x and y coordinates of all relative maximum points

 

 

C) Find the x and y coordinates of all relative minimum points.

 

 

D) Find the interval on which f is concave downward

 

Thanks in advance

Answer 1

f(x) =12x² / (3 - 4x)

 

 

Part A)

 

To find intervals where f is increasing, we need to perform the first derivative test.  In this test, we set the derivative of f equal to zero.  Use the quotient rule to derive f.

 

f'(x) = 0

 

[(24x(3 - 4x) - (12x²)(-4)] / (3 - 4x)² = 0

 

(72x - 96x² + 48x²) / (3 - 4x)² = 0

 

(72x - 48x²) / (3 - 4x)² = 0

 

 

Set the numerator equal to zero.

 

72x - 48x² = 0

 

24x(3 - 2x) = 0

 

24x = 0            and               3 - 2x = 0

 

x = 0                and                   -2x = -3

 

                                                    x = 1.5

 

 

The x values that we obtained are our critical points.  They are also the location of the possible minimum and maximum values of f(x).

 

Next, we use test points.  Evaluate the derivative using the following test points:

 

x = -1  ,   x = 1  ,  x = 2

 

 

Evaluate the following derivatives:

 

f'(-1)  ,  f'(1)  ,   f'(2)

 

 

If the derivative is positive, then f(x) is increasing.

 

 

Part B)

 

If the derivative of f(x) is positive before the critical point and negative after the critical point when evaluating the derivative using the test points, then f(x) is a maximum.

 

 

Part C)

 

If the derivative of f(x) is negative before the critical point and positive after the critical point when evaluating the derivative using the test points, then f(x) is a minimum.

 

 

Part D)

 

To find the concavities, we set the second derivative equal to zero.

 

d/dx[(72x - 48x²) / (3 - 2x)²] = 0

 

 

Solve for x the same way we did for first derivative test.  This time, the x values are our points of inflection.  Points of inflection are locations where the graph changes concavity.

 

Use the test points here too.

 

It is concave down if the second derivative is negative.