graph the function. give the minimum and maximum values over the specified interval. f(x)=√x + 1, 0≤x≤4 the…

Answer:

The minimum value is $1$. The maximum value is $3$.

Explanation:

Step1: Analyze the function's behavior

The function $y = f(x)=\sqrt{x}+1$ is an increasing - function for $x\geq0$ since the square - root function $y = \sqrt{x}$ is increasing for $x\geq0$.

Step2: Find the minimum value

When $x = 0$ (the left - hand endpoint of the interval $[0,4]$), we substitute $x = 0$ into the function $f(x)=\sqrt{x}+1$. Then $f(0)=\sqrt{0}+1=1$.

Step3: Find the maximum value

When $x = 4$ (the right - hand endpoint of the interval $[0,4]$), we substitute $x = 4$ into the function $f(x)=\sqrt{x}+1$. Then $f(4)=\sqrt{4}+1=2 + 1=3$.