Question

(3) f(x) = 2x² + x + 2; 0, 1

Explanation:

Step1: Identify the function and interval

We have the function \( f(x) = 2x^2 + x + 2 \) and the interval \([0, 1]\). We can analyze the behavior of the function on this interval, for example, to find its average rate of change or to check if it's increasing/decreasing. Let's find the average rate of change first. The formula for the average rate of change of a function \( y = f(x) \) over the interval \([a, b]\) is \( \frac{f(b)-f(a)}{b - a} \).

Step2: Calculate \( f(0) \) and \( f(1) \)

First, calculate \( f(0) \): substitute \( x = 0 \) into \( f(x) \), we get \( f(0)=2(0)^2+0 + 2=2 \).
Then, calculate \( f(1) \): substitute \( x = 1 \) into \( f(x) \), we get \( f(1)=2(1)^2+1 + 2=2 + 1+ 2=5 \).

Step3: Calculate the average rate of change

Using the formula for average rate of change, with \( a = 0 \), \( b = 1 \), we have \( \frac{f(1)-f(0)}{1 - 0}=\frac{5 - 2}{1}=3 \).
If we want to check if the function is increasing on \([0, 1]\), we can find its derivative. The derivative of \( f(x) \) is \( f'(x)=4x + 1 \). For \( x\in[0, 1] \), \( 4x+1\geq1>0 \), so the function is increasing on \([0, 1]\).

Answer:

If we consider the average rate of change, it is \( 3 \). If we consider the derivative analysis, the function is increasing on \([0, 1]\) with derivative \( f'(x)=4x + 1 \). (Depending on the specific question, but based on the given function and interval, these are common analyses.)