Chapter 8.4, Problem 110E

To determine

To graph: the given functions in same viewing window and identify the identical graphs.

Answer to Problem 110E

The graphs of (a) and (d) are the same.

Explanation of Solution

Given information:

Given functions are

  $\begin{array}{l}(a)f\left(x\right)={\left(1-\frac{x}{2}\right)}^4\\ (b)g\left(x\right)=1-2x+\frac{3}{2}{x}^2\\ (c)h\left(x\right)=1-2x+\frac{3}{2}{x}^2-\frac{1}{2}{x}^3\\ (d)p\left(x\right)=1-2x+\frac{3}{2}{x}^2-\frac{1}{2}{x}^3+\frac{1}{16}{x}^4\end{array}$

Calculation:

Use graphing software to graph the above4 functions. The graph of the 4 function are given by,

  

Therefore, the graphs of (a) and (d) are the same. One can also perform binomial expansion to check the answer. Let us consider the functions,

  $\begin{array}{l}f\left(x\right)={\left(1-\frac{x}{2}\right)}^4\\ p\left(x\right)=1-2x+\frac{3}{2}{x}^2-\frac{1}{2}{x}^3+\frac{1}{16}{x}^4\end{array}$

It is fact that,

  ${\left(x+y\right)}^4=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4$

Replace $x$ by 1 and $y$ by $-\frac{x}{2}$ ,

  $\begin{array}{c}{\left(1-\frac{x}{2}\right)}^4=1^4+4{\left(1\right)}^3\left(-\frac{x}{2}\right)+6{\left(1\right)}^2{\left(-\frac{x}{2}\right)}^2+4\left(1\right){\left(-\frac{x}{2}\right)}^3+{\left(-\frac{x}{2}\right)}^4\\ =1-2x+\frac{3}{2}{x}^2-\frac{1}{2}{x}^3+\frac{1}{16}{x}^4\\ =p\left(x\right)\end{array}$

Therefore, the graphs of (a) and (d) are the same.