Chapter 4.1, Problem 58PPS

a.

To determine

Make a table for functions if, $-4\le x\le 4$ .

Answer to Problem 58PPS

  $\begin{array}{cccccccccc}x& -4& -3& -2& -1& 0& 1& 2& 3& 4\\ f\left(x\right)& 40& 29& 20& 13& 8& 5& 4& 5& 8\\ g\left(x\right)& 88& 56& 32& 16& 8& 8& 16& 32& 56\end{array}$

Explanation of Solution

Given information:

  $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$

Calculation:

Consider the $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$ for $-4\le x\le 4$

Now calculate the value for functions $f\left(x\right)andg\left(x\right)$ for $-4\le x\le 4$

  $\begin{array}{l}putx=-4inf\left(x\right)andg\left(x\right)\\ f\left(x\right)=x^2-4x+8\\ f\left(-4\right)={\left(-4\right)}^2-4\left(-4\right)+8\\ =40\\ g\left(x\right)=4x^2-4x+8\\ g\left(-4\right)=4{\left(-4\right)}^2-4\left(-4\right)+8\\ =88\end{array}$

  $\begin{array}{l}putx=-3inf\left(x\right)andg\left(x\right)\\ f\left(-3\right)={\left(-3\right)}^2-4\left(-3\right)+8\\ =29\\ g\left(-3\right)=4{\left(-3\right)}^2-4\left(-3\right)+8\\ =56\end{array}$

  $\begin{array}{l}putx=-2inf\left(x\right)andg\left(x\right)\\ f\left(-2\right)={\left(-2\right)}^2-4\left(-2\right)+8\\ =20\\ g\left(-2\right)=4{\left(-2\right)}^2-4\left(-2\right)+8\\ =32\end{array}$

  $\begin{array}{l}putx=-1inf\left(x\right)andg\left(x\right)\\ f\left(-1\right)={\left(-1\right)}^2-4\left(-1\right)+8\\ =13\\ g\left(-1\right)=4{\left(-1\right)}^2-4\left(-1\right)+8\\ =16\end{array}$

  $\begin{array}{l}putx=0inf\left(x\right)andg\left(x\right)\\ f\left(0\right)={\left(0\right)}^2-4\left(0\right)+8\\ =8\\ g\left(0\right)=4{\left(0\right)}^2-4\left(0\right)+8\\ =8\end{array}$

  $\begin{array}{l}putx=1inf\left(x\right)andg\left(x\right)\\ f\left(1\right)={\left(1\right)}^2-4\left(1\right)+8\\ =5\\ g\left(1\right)=4{\left(1\right)}^2-4\left(1\right)+8\\ =8\end{array}$

  $\begin{array}{l}putx=2inf\left(x\right)andg\left(x\right)\\ f\left(2\right)={\left(2\right)}^2-4\left(2\right)+8\\ =4\\ g\left(2\right)=4{\left(2\right)}^2-4\left(2\right)+8\\ =16\end{array}$

  $\begin{array}{l}putx=3inf\left(x\right)andg\left(x\right)\\ f\left(3\right)={\left(3\right)}^2-4\left(3\right)+8\\ =5\\ g\left(3\right)=4{\left(3\right)}^2-4\left(3\right)+8\\ =32\end{array}$

  $putx=4inf\left(x\right)andg\left(x\right)$

  $\begin{array}{l}f\left(4\right)={\left(4\right)}^2-4\left(4\right)+8\\ =8\\ g\left(4\right)=4{\left(4\right)}^2-4\left(4\right)+8\\ =56\end{array}$

Now table for the functions,

  $\begin{array}{cccccccccc}x& -4& -3& -2& -1& 0& 1& 2& 3& 4\\ f\left(x\right)& 40& 29& 20& 13& 8& 5& 4& 5& 8\\ g\left(x\right)& 88& 56& 32& 16& 8& 8& 16& 32& 56\end{array}$

Hence, the function tabulated.

b.

To determine

Graph the functions $f\left(x\right)andg\left(x\right)$ .

Answer to Problem 58PPS

  

  

Explanation of Solution

Given information:

  $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$

Calculation:

Consider the $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$

Now graph $f\left(x\right)andg\left(x\right)$ for $-4\le x\le 4$

  

  

Hence, the graphs plotted.

c.

To determine

Find the difference in shape of graph of functions $f\left(x\right)andg\left(x\right)$ .

Answer to Problem 58PPS

Function $g\left(x\right)$ increases more rapidly than $f\left(x\right)$ .

Explanation of Solution

Given information:

  $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$

Calculation:

Consider the graph of $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$

Function $f\left(x\right)$ has minimum value $4$ at the $x$ -coordinate of the vertex $x=2$

Function $g\left(x\right)$ has minimum value $7$ at the $x$ -coordinate of the vertex $x=0.5$

Hence, function $g\left(x\right)$ increases more rapidly than $f\left(x\right)$ .

d.

To determine

Graph all three functions $f\left(x\right),g\left(x\right)andh\left(x\right)$ .

Answer to Problem 58PPS

  

Explanation of Solution

Given information:

  $f\left(x\right)=x^2-4x+8andg\left(x\right)=4x^2-4x+8$

Predict the appearance of graph $h\left(x\right)=0.25x^2-4x+8$ if $-10\le x\le 10$ .

Calculation:

Consider the $h\left(x\right)=0.25x^2-4x+8$ , $h\left(x\right)$ Function has positive $x$ -coefficient so, this quadratic function is in the form of parabola,

  $\begin{array}{l}x=-\frac{b}{2a}\\ =-\frac{-4}{2\left(0.25\right)}\\ =8\end{array}$

Now graph of three functions would be,