Question

Find the value of $h\left(-1\right),h\left(-2\right),h\left(\frac{1}{2}\right),h\left(x-1\right),h\left(\frac{1}{x}\right)$for the function $h\left(t\right)=t+\frac{1}{t}$.

Short answer

The value of function are

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

Step-by-step solution

Step 1. Evaluate h−1,h2,h12 for the given function.

From the given function,

$h\left(t\right)=t+\frac{1}{t}\mathrm{....}\left(1\right)$

Evaluate the value of function at $t=-1,2,\frac{1}{2}$gives,

role="math" localid="1644499246446" $$\begin{gathered} h\left(-1\right)=\left(-1\right)+\frac{1}{\left(-1\right)}\mathrm{....}\left(\text{Substitute}t=-1\text{in}\left(1\right)\right) \\ =-2 \\ h\left(2\right)=\left(-2\right)+\frac{1}{\left(-2\right)}\mathrm{....}\left(\text{Substitutet}=2\text{\hspace{0.17em}in}\left(1\right)\right) \\ =-\frac{5}{2} \\ h\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)+\frac{1}{\left(\frac{1}{2}\right)}\mathrm{....}\left(\text{Substitutet}=\frac{1}{2}\text{in}\left(1\right)\right) \\ =\frac{5}{2} \end{gathered}$$

Step 2. Evaluate hx−1,h1x for the given function.

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

role="math" localid="1644499418742" $$\begin{gathered} h\left(x-1\right)=\left(x-1\right)+\frac{1}{\left(x-1\right)}\mathrm{....}\left(\text{Substitute}t=x-1\text{in}\left(1\right)\right) \\ =\frac{x^2-2x+2}{x-1} \\ h\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)+\left(\frac{x}{1}\right)\mathrm{....}\left(\text{Substitutet}=\frac{1}{x}\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{1+x^2}{x} \end{gathered}$$

Step 3. Description of steps.

From the obtained values,

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