Chapter 9.6, Problem 36E

To determine

To calculate the solution of the given set of equations.

Answer to Problem 36E

The solution is $x\approx -0.41$ and $y\approx -2.51$

Explanation of Solution

Given information:

  $\begin{array}{l}y=-x^2-4x-4\\ y=-5^x-2\end{array}$

Formula used:

Approximation

Calculation:

Equation 1 is $y=-x^2-4x-4$

Equation 2 is $y=-5^x-2$

Graphing the two equations,

  

Concluding from the graph, the set of equation has two solution, which lines between $x=0$ and $x=-1$ ,

  $x=-3$ and $x=-4$ ,

Substituting equation 2 in equation 1,

  $-5^x-2=-x^2-4x-4$

Shifting the values,

  $-x^2-4x-4+5^x+2=0$

On solving,

  $-x^2-4x+5^x-2=0$

Since, this equations cannot be solved algebraically, let,

  $f(x)=-x^2-4x+5^x-2$

Evaluating f(x) between $x=0$ and $x=-1$ ,

For f(-0.4), substituting the value,

  $f(-0.4)=-{(0.4)}^2-4(-0.4)+5^{-0.4}-2$

On solving,

  $f(-0.4)=-0.03$

For f(0.5), substituting the value,

  $f(-0.5)=-{(0.5)}^2-4(-0.5)+5^{-0.5}-2$

On solving,

  $f(-0.5)=0.20$

Since, f(-0.4)<0 and f(-0.5)>0, the zero is between -0.4 and -0.5

Since f(-0.4) is closer to zero than f(-0.5), so increases the guess and evaluate f(-0.4)

For f(-0.41) substituting the values,

  $f(-0.41)=-{(0.41)}^2-4(-0.41)+5^{-0.41}-2$

On solving,

  $f(-0.41)=0.186$

For f(-0.42) substituting the values,

  $f(-0.42)=-{(0.42)}^2-4(-0.42)+5^{-0.42}-2$

On solving,

  $f(-0.42)=0.2$

Because, f(-0.41) is the closest to 0

Therefore, $x\approx -0.41$

For y, substituting the values,

  $y=-{(-0.41)}^2-4(-0.41)-4$

On solving,

  $y\approx -2.51$

Hence, the solution is $x\approx -0.41$ and $y\approx -2.51$