Question

given the function
$f(x)=\

$$\begin{cases}8x - 5&x < 0\\\\\\sqrt{8x^2 + 25}&x \\geq 0\\end{cases}$$

$
calculate the following values round answers to three decimals as necessary:
$f(-1) = $
$f(0) = $
$f(2) = $
question help: message instructor

Explanation:

Step1: Evaluate $f(-1)$ (use $x<0$ rule)

Since $-1 < 0$, substitute $x=-1$ into $8x-5$:

$$\begin{align*} f(-1) &= 8(-1) - 5 \\ &= -8 - 5 \\ &= -13 \end{align*}$$

Step2: Evaluate $f(0)$ (use $x\geq0$ rule)

Since $0 \geq 0$, substitute $x=0$ into $\sqrt{8x^2+25}$:

$$\begin{align*} f(0) &= \sqrt{8(0)^2 + 25} \\ &= \sqrt{0 + 25} \\ &= \sqrt{25} \\ &= 5 \end{align*}$$

Step3: Evaluate $f(2)$ (use $x\geq0$ rule)

Since $2 \geq 0$, substitute $x=2$ into $\sqrt{8x^2+25}$:

$$\begin{align*} f(2) &= \sqrt{8(2)^2 + 25} \\ &= \sqrt{8(4) + 25} \\ &= \sqrt{32 + 25} \\ &= \sqrt{57} \\ &\approx 7.550 \end{align*}$$

Answer:

$f(-1) = -13$
$f(0) = 5$
$f(2) = 7.550$