Intermediate Value Theorem Solution 2

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SOLUTION 2: We are given the equation

$e^{-x}=4+x^3 \ \ \ \ \longrightarrow \ \ \ \ e^{-x}-x^3-4=0$ and the interval

$[-2, -1]$ . Let function

$f(x)=e^{-x}-x^3-4 \ \ \ \ and \ choose \ \ \ \ m=0$ This function is continuous for all values of

$x$ since it is the DIFFERENCE of continuous functions; it is well-known that

$e^{-x}$ is continuous for all values of

$x$ and

$x^3+4$ is continuous for all values of

$x$ since it is a polynomial. (Please note that the graph of the function is not necessary for a valid proof, but the graph will help us understand how to use the Intermediate Value Theorem. On many subsequent problems, we will solve the problem without using the "luxury" of a graph.)

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Note that

$f(-2)= e^{-(-2)}-(-2)^3-4 = e^2+4 \approx 11.4 > 0 \ \ \ \ and \ \ \ \ f(-1)= e^{-(-1)}-(-1)^3-4 = e-3 \approx -0.3 <0$
so that

$f(-1) \approx -0.3 < m < 11.4 \approx f(-2)$
i.e.,

$m=0$ is between

$f(-1)$ and

$f(-2)$ .

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

$c$ in the interval

$[-2, -1]$ which satisfies

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

$e^{-c}-c^3-4=0$ and the equation is solvable.

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