Question

Find the value of $f\left(-3\right),f\left(3\right),f\left(0\right),f\left(\frac{1}{2}\right)$for the function $f\left(x\right)=x^2-6$.

Short answer

The value of function are $f\left(-3\right)=3,f\left(3\right)=3,f\left(0\right)=-6\text{\hspace{0.17em}and}f\left(\frac{1}{2}\right)=-\frac{23}{4}$.

Step-by-step solution

Step 1. Evaluate f−3,f3 for the function.

From the given function,

$f\left(x\right)=x^2-6....\left(1\right)$

Evaluate the value of function at $x=-3,3$gives,

role="math" localid="1644493697695" $$\begin{gathered} f\left(-3\right)={\left(-3\right)}^2-6....\left(\text{Substitute}x=-3\text{in}\left(1\right)\right) \\ =9-6 \\ =3 \\ f\left(3\right)={\left(3\right)}^2-6....\left(\text{Substitute}x=3\text{\hspace{0.17em}in}\left(1\right)\right) \\ =3 \end{gathered}$$

Step 2. Evaluate f0,f12 for the given function.

Similarly, evaluate the value of function (1) at $x=0,\frac{1}{2}$gives,

role="math" localid="1644494007341" $$\begin{gathered} f\left(0\right)={\left(0\right)}^2-6....\left(\text{Substitute}x=0\text{in}\left(1\right)\right) \\ =-6 \\ f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-6....\left(\text{Substitute}x=\frac{1}{2}\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{1}{4}-6 \\ =-\frac{23}{4} \end{gathered}$$

Step 3. Description of steps.

From the obtained values $f\left(-3\right)=3,f\left(3\right)=3,f\left(0\right)=-6\text{\hspace{0.17em}and}f\left(\frac{1}{2}\right)=-\frac{23}{4}$.