Question

Find the value of $h\left(2\right),h\left(-2\right),h\left(a\right),\left(-x\right),h\left(a-2\right),h\left(\sqrt{x}\right)$for the function $h\left(x\right)=\frac{x^2+4}{5}$.

Short answer

The value of function are

$h\left(2\right)=\frac{8}{5},h\left(-2\right)=\frac{8}{5},h\left(a\right)=\frac{a^2+4}{5},h\left(-x\right)=\frac{x^2+4}{5},h\left(a-2\right)=\frac{a^2-4a+8}{5},h\left(\sqrt{x}\right)=\frac{x+4}{5}$.

Step-by-step solution

Step 1. Evaluate h2,h−2,ha for the given function.

From the given function,

$h\left(x\right)=\frac{x^2+4}{5}\mathrm{....}\left(1\right)$

Evaluate the value of function at $x=2,-2,a$gives,

role="math" localid="1644497071786" $$\begin{gathered} h\left(2\right)=\frac{{\left(2\right)}^2+4}{5}\mathrm{....}\left(\text{Substitute}x=2\text{in}\left(1\right)\right) \\ =\frac{8}{5} \\ h\left(-2\right)=\frac{{\left(-2\right)}^2+4}{5}\mathrm{....}\left(\text{Substitute}x=-2\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{8}{5} \\ h\left(a\right)=\frac{{\left(a\right)}^2+4}{5}\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =\frac{a^2+4}{5} \end{gathered}$$

Step 2. Evaluate −x,ha−2,hx for the given function.

Similarly, evaluate the value of function (1) at $x=-x,a-2,\sqrt{x}$gives,

role="math" localid="1644497403252" $$\begin{gathered} h\left(-x\right)=\frac{{\left(-x\right)}^2+4}{5}\mathrm{....}\left(\text{Substitute}x=-x\text{in}\left(1\right)\right) \\ =\frac{x^2+4}{5} \\ h\left(a-2\right)=\frac{{\left(a-2\right)}^2+4}{5}\mathrm{....}\left(\text{Substitute}x=a-2\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{a^2-4a+8}{5} \\ h\left(\sqrt{x}\right)=\frac{{\left(\sqrt{x}\right)}^2+4}{5}....\left(\text{Substitute}x=\sqrt{x}\text{in}\left(1\right)\right) \\ =\frac{x+4}{5} \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

From the obtained values,

$h\left(2\right)=\frac{8}{5},h\left(-2\right)=\frac{8}{5},h\left(a\right)=\frac{a^2+4}{5},h\left(-x\right)=\frac{x^2+4}{5},h\left(a-2\right)=\frac{a^2-4a+8}{5},h\left(\sqrt{x}\right)=\frac{x+4}{5}$.