Chapter 11, Problem 3STP

To determine

To evaluate: the point that lies on the graph of $\left(gof\right)\left(x\right)$ where $f\left(x\right)=2x^2$ and $g\left(x\right)=\sqrt{x+1}$ .

Answer to Problem 3STP

Option C and D are correct.

Explanation of Solution

Given:

  $f\left(x\right)=2x^2$

  $g\left(x\right)=\sqrt{x+1}$

In this, finding the $\left(gof\right)\left(x\right)$ ,

  $\begin{array}{c}g\left(f\left(x\right)\right)=\sqrt{x+1}\\ =\sqrt{2x^2+1}\end{array}$

Further,

Checking (−1, 2)

  $\begin{array}{c}g\left(f\left(-1\right)\right)=\sqrt{2x^2+1}\\ =\sqrt{2{\left(-1\right)}^2+1}\\ =\sqrt{3}\end{array}$

So, this point does not lie on the graph.

Checking (0, 2)

  $\begin{array}{c}g\left(f\left(0\right)\right)=\sqrt{2x^2+1}\\ =\sqrt{2{\left(0\right)}^2+1}\\ =\sqrt{1}\\ =1\end{array}$

So, this point does not lie on the graph.

Checking (1, 2)

  $\begin{array}{c}g\left(f\left(0\right)\right)=\sqrt{2x^2+1}\\ =\sqrt{2{\left(0\right)}^2+1}\\ =\sqrt{1}\\ =1\end{array}$

So, this point does lie on the graph.

Checking (2, 3)

  $\begin{array}{c}g\left(f\left(2\right)\right)=\sqrt{2x^2+1}\\ =\sqrt{2{\left(2\right)}^2+1}\\ =\sqrt{9}\\ =3\end{array}$

So, this point does lie on the graph.

Hence option C and D are correct.