Question

suppose that the function f is defined, for all real numbers, as follows.
$f(x)=\

$$\begin{cases} -\\dfrac{1}{4}x^2 + 4 & \\text{if } x\ eq 2 \\\\ 1 & \\text{if } x = 2 \\end{cases}$$

$
find $f(-5)$, $f(2)$, and $f(4)$.
$f(-5) = \square$
$f(2) = \square$
$f(4) = \square$

Explanation:

Step1: Calculate $f(-5)$ (use $x

eq2$ rule)
Substitute $x=-5$ into $-\frac{1}{4}x^2 + 4$:

$$\begin{align*} f(-5)&=-\frac{1}{4}(-5)^2 + 4\\ &=-\frac{1}{4}(25) + 4\\ &=-\frac{25}{4} + \frac{16}{4}\\ &=-\frac{9}{4} \end{align*}$$

Step2: Calculate $f(2)$ (use $x=2$ rule)

Use the given value for $x=2$:
$f(2)=1$

Step3: Calculate $f(4)$ (use $x

eq2$ rule)
Substitute $x=4$ into $-\frac{1}{4}x^2 + 4$:

$$\begin{align*} f(4)&=-\frac{1}{4}(4)^2 + 4\\ &=-\frac{1}{4}(16) + 4\\ &=-4 + 4\\ &=0 \end{align*}$$

Answer:

$f(-5)=-\frac{9}{4}$
$f(2)=1$
$f(4)=0$