Question

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

Short answer

The value of function are

$f\left(0\right)=0,f\left(3\right)=15,f\left(-3\right)=3,f\left(a\right)=a^2+2a,f\left(-x\right)=x^2-2x,f\left(\frac{1}{a}\right)=\frac{1}{a^2}+\frac{2}{a}$.

Step-by-step solution

Step 1. Evaluate f0,f3,f−3 for the given function.

From the given function,

$f\left(x\right)=x^2+2x\mathrm{....}\left(1\right)$

Evaluate the value of function at role="math" localid="1644498105795" $x=0,3,-3$gives,

role="math" localid="1644498139041" $$\begin{gathered} f\left(0\right)={\left(0\right)}^2+2\left(0\right)\mathrm{....}\left(\text{Substitute}x=0\text{in}\left(1\right)\right) \\ =0 \\ f\left(3\right)={\left(3\right)}^2+2\left(3\right)\mathrm{....}\left(\text{Substitute}x=3\text{\hspace{0.17em}in}\left(1\right)\right) \\ =15 \\ f\left(-3\right)={\left(-3\right)}^2+2\left(-3\right)\mathrm{....}\left(\text{Substitute}x=-3\text{in}\left(1\right)\right) \\ =3 \end{gathered}$$

Step 2. Evaluate fa,f−x,f1a the given function.

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

role="math" localid="1644498430327" $$\begin{gathered} f\left(a\right)={\left(a\right)}^2+2\left(a\right)\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =a^2+2a \\ f\left(-x\right)={\left(-x\right)}^2+2\left(-x\right)\mathrm{....}\left(\text{Substitute}x=-x\text{\hspace{0.17em}in}\left(1\right)\right) \\ =x^2-2x \\ f\left(\frac{1}{a}\right)={\left(\frac{1}{a}\right)}^2+2\left(\frac{1}{a}\right)\mathrm{....}\left(\text{Substitute}x=\frac{1}{a}\text{in}\left(1\right)\right) \\ =\frac{1}{a^2}+\frac{2}{a} \end{gathered}$$

Step 3. Description of steps.

From the obtained values,

$f\left(0\right)=0,f\left(3\right)=15,f\left(-3\right)=3,f\left(a\right)=a^2+2a,f\left(-x\right)=x^2-2x,f\left(\frac{1}{a}\right)=\frac{1}{a^2}+\frac{2}{a}$.