Question

For the function $h\left(x\right)=2x+1$, evaluate the function.

1. $h\left(k^2\right)$
   
2. $h\left(x+1\right)$
   
3. $h\left(x\right)+h\left(1\right)$
   

Short answer

Part (a)$h\left(k^2\right)=2k^2+1$

Part (b) $h\left(x+1\right)=2x+3$

Part (c) $h\left(x\right)+h\left(1\right)=2x+4$

Step-by-step solution

Part (a) Step 1. 

Substitute $x=k^2$in localid="1644749461383" $h\left(x\right)=2x+1$,

$\begin{array}{rcl}h\left(k^2\right)& =& 2\left(k^2\right)+1\\ & =& 2k^2+1\end{array}$

Part (b) Step 1. 

Substitute $x+1$for $x$in localid="1644749540514" $h\left(x\right)=2x+1$,

$\begin{array}{rcl}h\left(x+1\right)& =& 2\left(x+1\right)+1\\ & =& 2x+2+1\\ & =& 2x+3\end{array}$

Part (c) Step 1. 

$\begin{array}{rcl}h\left(x\right)+h\left(1\right)& =& \left(2x+1\right)+\left(2\left(1\right)+1\right)\\ & =& 2x+1+\left(2+1\right)\\ & =& 2x+1+3\\ & =& 2x+4\end{array}$