Question

Part 1 of 3Points: 0 of 1For the function below, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is decreasing.f(x)=2.8+1.6x-0.3x²(a) The critical number(s) is/are.(Type an integer or a simplified fraction. Use a comma to separate answers as needed.)

Answer 1

Answer

(a) The critical number is 8/3

(b) Increasing on (-∞, 8/3)

(c) Decreasing on (8/3, ∞)

Step-by-step explanation

(a) Given the function:

$f(x)=2.8+1.6x-0.3x^2$

To find the critical number, first, we need to compute the derivative of f(x), as follows:

$\begin{gathered} \frac{d}{dx}f(x)=\frac{d}{dx}(2.8+1.6x-0.3x^2) \\ f^{\prime}(x)=\frac{d}{dx}2.8+\frac{d}{dx}1.6x-\frac{d}{dx}0.3x^2 \\ f^{\prime}(x)=0+1.6\frac{d}{dx}x-0.3\frac{d}{dx}x^2 \\ f^{\prime}(x)=1.6-0.3\cdot2x \\ f^{\prime}(x)=1.6-0.6x \end{gathered}$

The critical numbers are those numbers where f'(x) is equal to zero. In this case:

$\begin{gathered} 0=1.6-0.6x \\ 0-1.6=1.6-0.6x-1.6 \\ -1.6=-0.6x \\ \frac{-1.6}{-0.6}=\frac{-0.6x}{-0.6} \\ \frac{8}{3}=x \end{gathered}$

(b) Now, we need to evaluate f'(x) in the interval of values less than the critical number and the values greater than the critical number. Taking for example x = 2 (2 is less than 8/3), we get:

$\begin{gathered} f^{\prime}(x)=1.6-0.6x \\ f^{\prime}(2)=1.6-0.6\cdot2 \\ f^{\prime}(2)=0.4 \end{gathered}$

Given that f'(x) is positive, then f(x) is increasing in the interval (-∞, 8/3)

(c) Taking now for example x = 3 (3 is greater than 8/3), we get:

$\begin{gathered} f^{\prime}(x)=1.6-0.6x \\ f^{\prime}(3)=1.6-0.6\cdot3 \\ f^{\prime}(3)=-0.2 \end{gathered}$

Given that f'(x) is negative, then f(x) is decreasing in the interval (8/3, ∞)