Question

3 numeric 1 point
given:
$f(x)=\

$$\begin{cases}x^{2}+1; & x\\leq -1 \\\\ x + 4; & -1< x<2 \\\\ 5; & x\\geq 2\\end{cases}$$

$
find: $-2f(-4)+f(7)-f(0)$
answer

Explanation:

Step1: Calculate $f(-4)$

Since $-4 \leq -1$, use $f(x)=x^2+1$.
$f(-4)=(-4)^2+1=16+1=17$

Step2: Calculate $f(7)$

Since $7 \geq 2$, use $f(x)=5$.
$f(7)=5$

Step3: Calculate $f(0)$

Since $-1<0<2$, use $f(x)=x+4$.
$f(0)=0+4=4$

Step4: Substitute into the expression

$-2f(-4)+f(7)-f(0)=-2(17)+5-4$

Step5: Compute the final value

$-34+5-4=-33$

Answer:

-33