Chapter 4.4, Problem 74E

To determine

To show:The shrinking of the graph of ‘f’ horizontally the function $f\left(x\right)=x^2$ .

  $f\left(2x\right)=4f\left(x\right)$

To use the identities $e^{2+x}=e^2{e}^x$ and $\ln \left(2x\right)=\ln 2+\ln x$ to show for $g\left(x\right)=e^x$ and for $h\left(x\right)=\ln x$

Answer to Problem 74E

The $f\left(x\right)$ to $$$f\left(2x\right)$ curves to shrink, also it starches vertically.

Horizontal shrink, $h\left(2x\right)=\ln 2x=\ln 2+\ln x=$ vertically shift.

Explanation of Solution

Given information:

The expression:

The shrinking of the graph of ‘f’ horizontally the function $f\left(x\right)=x^2$ .

  $f\left(2x\right)=4f\left(x\right)$

To use the identities $e^{2+x}=e^2{e}^x$ and $\ln \left(2x\right)=\ln 2+\ln x$ to show for $g\left(x\right)=e^x$ and for $h\left(x\right)=\ln x$

Formula used:

Using the laws of logarithm:

  $e^{2+x}=e^2{e}^x$ and $\ln \left(2x\right)=\ln 2+\ln x$

Consider the shrinking of the graph of ‘f’ horizontally the function $f\left(x\right)=x^2$ .

  $f\left(2x\right)=4f\left(x\right)$

Now the function $f\left(x\right)=x^2$ .

  $\begin{array}{c}f\left(x\right)=x^2\\ f\left(2x\right)={\left(2x\right)}^2\\ =4x^2\\ =4f\left(x\right)\end{array}$

Now the graph of the functions $f\left(x\right)=x^2,f\left(x\right)=4x^2.$ shown below,

  

Here see in the above graph the red curve shows the graph of $f\left(x\right)=x^2$ and the blue curve shows the graph for $f\left(x\right)=4x^2.$

From here observe that the value of function of $x$ increases then the graph take more vertical shape.

Now for the function for,

  $g\left(x\right)=e^x$

Horizontal shift, $g\left(x+2\right)=e^{x+2}=e^2{e}^x=$ vertically stretch.

  $h\left(x\right)=\ln x$

Horizontal shrink of the function is:

  $\begin{array}{c}h\left(2x\right)=\ln 2x\\ =\ln 2+\ln x\\ =\ln 2+h\left(x\right)\end{array}$

Are all equal to vertical shift.

The horizontal stretching is the same as a vertically shift.