Chapter 9.6, Problem 33E

To determine

To calculate the solution of the given set of equations.

Answer to Problem 33E

The solution is $x\approx 2.35$ and $y\approx 13.22$

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^2+2x+3\\ y=3^x\end{array}$

Formula used:

Approximation

Calculation:

Equation 1 is $y=x^2+2x+3$

Equation 2 is $y=3^x$

Graphing the two equations,

  

Concluding from the graph, the set of equation has one solution, which lines between $x=2$ and $x=3$

Substituting equation 2 in equation 1,

  $3^x=x^2+2x+3$

Shifting the values,

  $x^2+2x+3-3^x=0$

Since, this equations cannot be solved algebraically, let,

  $f(x)=x^2+2x+3-3^x$

Evaluating f(x) between $x=2$ and $x=3$ ,

For f(2.3), substituting the value,

  $f(2.3)={(2.3)}^2+2(2.3)+3-3^{2.3}$

On solving,

  $f(2.3)=0.37$

For f(2.4), substituting the value,

  $f(2.4)={(2.4)}^2+2(2.4)+3-3^{2.4}$

On solving,

  $f(2.4)=-0.407$

Since, f(2.3)>0 and f(2.4)<0, the zero is between 2.3 and 2.4

Since f(2.3) is closer to zero than f(2.4), so increases the guess and evaluate f(2.31)

For f(2.31) substituting the values,

  $f(2.31)={(2.31)}^2+2(2.31)+3-3^{2.31}$

On solving,

  $f(2.31)=0.30$

For f(2.32) substituting the values,

  $f(2.32)={(2.32)}^2+2(2.32)+3-3^{2.32}$

On solving,

  $f(2.32)=0.2$

For f(2.33) substituting the values,

  $f(2.33)={(2.33)}^2+2(2.33)+3-3^{2.33}$

On solving,

  $f(2.33)=0.16$

For f(2.34) substituting the values,

  $f(2.34)={(2.34)}^2+2(2.34)+3-3^{2.34}$

On solving,

  $f(2.34)=0.080$

For f(2.35) substituting the values,

  $f(2.35)={(2.35)}^2+2(2.35)+3-3^{2.35}$

On solving,

  $f(2.35)=0.002$

For f(2.36) substituting the values,

  $f(2.36)={(2.36)}^2+2(2.36)+3-3^{2.36}$

On solving,

  $f(2.36)=-0.07$

Because, f(2.35) is the closest to 0

Therefore, $x\approx 2.35$

For y, substituting the values,

  $y=3^{2.35}$

On solving,

  $y\approx 13.22$

Hence, the solution is $x\approx 2.35$ and $y\approx 13.22$