Intermediate Value Theorem Solution 9

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SOLUTION 9: We are given the equation

$x e^x = x^2-1 \ \ \ \ \longrightarrow \ \ \ \ x e^x - x^2+1=0$
Let function

$f(x)= xe^x - x^2+1 \ \ \ \ and \ choose \ \ \ \ m=0$
This function is continuous for all values of

$x$ since it is the SUM of continuous functions. We have the continuous polynomial function

$1-x^2$ . And the function

$xe^x$ is continuous for all values of

$x$ since it is the PRODUCT of

$x$ , a polynomial, and

$e^x$ , a well-known continuous function. We now need to search for an appropriate interval satisfying the assumptions of the Intermediate Value Theorem. By trial and error, we have that

$f(0)= (0)e^0 - (0)^2+1= 1 > 0 \ \ \ \ and \ \ \ \ f(-2)= (-2)e^{-2} - (-2)^2+1= \displaystyle{ -2 \over e^2 }-3 \approx -3.27 <0$
so that

$f(-2) \approx -3.27 < m < 1 = f(0)$
i.e.,

$m=0$ is between

$f(-2)$ and

$f(0)$ .

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number

$c$ in the interval

$[-2, 0]$ which satisfies

$f(c)=m$ i.e.,

$ce^c - c^2+1 =0$ and the equation is solvable.

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