Afunction is given. (a) find all the local maximum and minimum values of the function and the value of x at which each occurs. (b) find the intervals on which the function is increasing and on which the function is decreasing. state each answer correct to two decimal places. u (x) = 4 (x3 - x)

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

Step-by-step explanation:

To answer all of the questions we must obtain the derivative of the fucntion:

If

U(x) = 4(x^3 - x)

then

U'(x) = 4(3x^2 - 1)

U''(x) = 4(3*2 x) = 24 x

U'''(x) = 24

The local maxima and minima of the function U(x) can be found when U'(x) = 0

this occurs when :

3x^2 = 1

that is:

x = ±1/√3

We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as Second derivative test), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.

Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }

The first derivative will be positive when:

3x^2 > 1

|x| > 1/√3  -->  x > 1/√3   and    x < -1/√3

The first derivatice will be negative when:

3x^2 < 1

|x| < 1/√3  -->  x < 1/√3   and    x > -1/√3

Therefore the intervals are:

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum

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