Question

Which statement about the function is true? The function is increasing for all real values of x where x < 0. The function is increasing for all real values of x where x < –1 and where x > 4. The function is decreasing for all real values of x where –1 < x < 4. The function is decreasing for all real values of x where x < 1.5.

Answer 1

Answer:

Option D is correct.

The function is decreasing for all real values of x where x < 1.5.

Step-by-step explanation:

The function is given as

f(x) = (x – 4)(x + 1)

It can be simplified by expansion to give

f(x) = x² + x - 4x - 4 = x² - 3x - 4

To check which regions where the function is increasing, decreasing or constant, there are two ways to go about it.

The first method is the Calculus analysis.

In calculus analysis, the sign on the derivative of the function shows what is happening to the function in that interval.

Let the derivative of the function be f'(x)

- When f'(x) > 0, that is positive, the function is said to be increasing.

- When f'(x) < 0, that is negative, the function is said to be decreasing.

- When f'(x) = 0, that is equal to 0, the function is said to be constant.

The second method is to plot and examine the graph of the function. It becomes straight forward to just check the graph and see which regions of x where the function is increasing, decreasing or constant.

Using the first method first,

f(x) =  x² - 3x - 4

f'(x) = 2x - 3

This function would be increasing when f'(x) > 0

2x - 3 > 0

2x > 3

x > 1.5.

This means the function is increasing at values of x greater tthan 1.5.

And the function would be decreasing when f'(x) < 0

2x - 3 < 0

2x < 3

x < 1.5

This means that the value of x is decreasing at values of x less than 1.5.

Hence, option D is correct.

To check the other options if they are correct using this method, we just insert the values of x given into the expression obtained for f'(x) and check the sign to see if the function is increasing or decreasing at those points.

Option A: The function is increasing for all real values of x where x < 0

Picking a random value of x which is less than 0. x = -1.

f'(x) = 2x - 3 = 2(-1) - 3 = -5 (negative; indicates that the function is decreasing at this point)

Hence, this option is incorrect.

Option B: The function is increasing for all real values of x where x < –1 and where x > 4.

x = -2 and x = 5 will be tested.

f'(x) = 2x - 3 = 2(-2) - 3 = -7 (negative; indicates that the function is decreasing at this point)

f'(x) = 2x - 3 = 2(5) - 3 = 7 (positive; indicates that the function is increasing at this point)

Hence, only one half of the statement is true, so, the full statement can be adjuged to be false then.

Option C: The function is decreasing for all real values of x where –1 < x < 4.

x = 0 and x = 2

f'(x) = 2x - 3 = 2(0) - 3 = -3 (negative; indicates that the function is decreasing at this point)

f'(x) = 2x - 3 = 2(2) - 3 = 1 (positive; indicates that the function is increasing at this point)

Hence, it is evident that below 1.5, the function really is decreasing, but beyond that, the function starts to increase. This statement is false too.

Now using the graphical method, the graph of this function is attached to this solution. And from the graph, all the results of the Calculus analysis from above is evident just by looking at the graph.

The function decreases at values of x below x=1.5 and automatically starts to increase at values of x greater than x=1.5.

I have now exhaustively shown that only option D is correct for this question.

Hope this Helps!!!