Question Statement

Prove that the equation $\frac{x^2+1}{x+3}+\frac{x^4+1}{x-4}=0$ has a solution in the interval $(-3,4)$.

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Background and Explanation

This problem applies the Intermediate Value Theorem (IVT), which states that if a continuous function changes sign over an interval (from positive to negative or vice versa), it must equal zero at some point within that interval. The function has vertical asymptotes at $x=-3$ and $x=4$, but remains continuous everywhere on the open interval $(-3,4)$.

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Solution

Let us define the function:

$f(x)=\frac{x^2+1}{x+3}+\frac{x^4+1}{x-4}$

First, note that $f(x)$ is continuous on the open interval $(-3,4)$ because the denominators $(x+3)$ and $(x-4)$ are non-zero for all $x\in (-3,4)$.

Evaluating near $x=-3$

Since the interval $(-3,4)$ is open, the endpoint $-3$ is not included. We take an arbitrary value close to $-3$ from the right side. Let $x=-2.999$:

$\begin{array}{rl}f(-2.999)& =\frac{(-2.999{)}^2+1}{-2.999+3}+\frac{(-2.999{)}^4+1}{-2.999-4}\\ & =\frac{8.994001+1}{0.001}+\frac{80.89205399+1}{-6.999}\\ & =9994.001-11.70053636\\ & =9982.300464\\ & >0\end{array}$

Therefore, $f(-2.999)>0$.

Evaluating near $x=4$

Similarly, the endpoint $4$ is not included. We take a value closest to $4$ from the left side. Let $x=3.999$:

$\begin{array}{rl}f(3.999)& =\frac{(3.999{)}^2+1}{3.999+3}+\frac{(3.999{)}^4+1}{3.999-4}\\ & =\frac{15.992001+1}{6.999}+\frac{255.744096+1}{-0.001}\\ & =\frac{16.992001}{6.999}-\frac{256.744096}{0.001}\\ & =2.427775539-256744.096\\ & =-256741.6682\\ & <0\end{array}$

Therefore, $f(3.999)<0$.

Conclusion

We have established that:

• $f(-2.999)>0$ (positive value near $-3$)
• $f(3.999)<0$ (negative value near $4$)

Since $f(x)$ is continuous on the closed interval $[-2.999,3.999]$ (which is contained within $(-3,4)$), and the function values have opposite signs at the endpoints, by the Intermediate Value Theorem there must exist at least one value $c\in (-2.999,3.999)$ such that $f(c)=0$.

This shows that the given equation has a solution in the interval $(-3,4)$.

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Key Formulas or Methods Used

• Intermediate Value Theorem (IVT): If $f$ is continuous on $[a,b]$ and $f(a)\cdot f(b)<0$ (opposite signs), then there exists at least one $c\in (a,b)$ where $f(c)=0$
• Continuity of Rational Functions: A rational function is continuous everywhere on its domain (where the denominator is non-zero)
• Sign Analysis: Evaluating limits numerically by approaching asymptotes from within the domain

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Summary of Steps

1. Define the function $f(x)=\frac{x^2+1}{x+3}+\frac{x^4+1}{x-4}$ and confirm continuity on $(-3,4)$
2. Evaluate at $x=-2.999$ (approaching $-3$ from right): calculate $f(-2.999)=9982.300464>0$
3. Evaluate at $x=3.999$ (approaching $4$ from left): calculate $f(3.999)=-256741.6682<0$
4. Apply IVT: Since $f$ is continuous and changes sign between $-2.999$ and $3.999$, a root exists in $(-3,4)$