Question

6.2 symmetry and piecewise - defined functions
question 6 of 15 (4 points) | question attempt: 1 of unlimited
$f(x)=\

$$\begin{cases}-5x + 7&\\text{for }x < - 1\\\\x^{2}+2&\\text{for }-1\\leq x < 5\\\\6&\\text{for }x\\geq5\\end{cases}$$

$
part 1 of 5
$f(0)=\boxed{2}$
part 2 of 5
$f(-2)=\boxed{17}$
part 3 of 5
$f(-1)=\square$

Explanation:

Step1: Match x=-1 to correct piece

The value $x=-1$ satisfies $-1 \leq x < 5$, so use $f(x)=x^2+2$.

Step2: Substitute x=-1 into the function

$f(-1)=(-1)^2 + 2$

Step3: Calculate the result

$(-1)^2 + 2 = 1 + 2 = 3$

Answer:

3