Chapter 7.6, Problem 31P

Open-Ended Find the value of each of the functions a) $f(x)=2x^2$ and b) $f(x)=2\cdot 2^x$ for $x=5.$

Write another quadratic function and another exponential function with a base of two whose values at $x=5$ are between the values you found for functions a and b.

To determine

To calculate:Find the value of the function $f\left(x\right)=2x^2$ and $f\left(x\right)=2*2^x$ for $x=5.$ Write another quadratic function and another exponential function with base of two whose values at $x=5$ are between the values you found for above functions.

Answer to Problem 31P

The value of the function $f\left(x\right)=2x^2$ is 50 and the value of the function $f\left(x\right)=2*2^x$ is 64. The new functions are $f\left(x\right)=2.1*x^2$ and $f\left(x\right)=1.75*2^x$ .

Explanation of Solution

Given information:The given functions are $f\left(x\right)=2x^2$ and $f\left(x\right)=2*2^x$ .

Calculation:First we will find the value of both the functions at $x=5$ ,

For the function $f\left(x\right)=2x^2$ , putting $x=5$ , we get

  $\begin{array}{l}f\left(5\right)=2{\left(5\right)}^2\\ f\left(5\right)=2*25\\ f\left(5\right)=50\end{array}$

For the function $f\left(x\right)=2*2^x$ , putting $x=5$ , we get

  $\begin{array}{l}f\left(5\right)=2{\left(2\right)}^5\\ f\left(5\right)=2*32\\ f\left(5\right)=64\end{array}$

Now we write a quadratic equation whose value lie between 50 and 64

  $f\left(x\right)=2.1*x^2$

and $f\left(5\right)=2.1*5^2=52.5$

Now we write a exponential function whose value lie between 50 and 64

  $f\left(x\right)=1.75*2^x$

and $f\left(5\right)=1.75*2^5=56$

Here both the value 52.5 and 56 lies between 50 and 64.