Show how to use the Intermediate Value Theorem to show that the equation $\displaystyle \frac{{6}^{{x}}}{{{1}+{x}^{{6}}}}=\frac{{4}}{{3}}$ has a solution between 1 and 4.

Let $\displaystyle {f{{\left({x}\right)}}}=\frac{{6}^{{x}}}{{{1}+{x}^{{6}}}}$. In order for the Intermediate Value Theorem to apply we must first check that $\displaystyle {f}$ is Select an answer solvable a rational function a polynomial differentiable continuous on the interval $\displaystyle {\left[{1},{4}\right]}$. You should verify this and be able to explain why it is the case.

We check the value of $\displaystyle {f}$ at the left endpoint of the interval $\displaystyle {\left[{1},{4}\right]}$:

$\displaystyle {f}$() = .

And we check the value of $\displaystyle {f}$ at the right endpoint of the interval $\displaystyle {\left[{1},{4}\right]}$:

$\displaystyle {f}$() = Preview Question 6 Part 5 of 8   .

The Intermediate Value Theorem guarantees a solution to $\displaystyle {f{{\left({x}\right)}}}=\frac{{4}}{{3}}$ in the interval $\displaystyle {\left({1},{4}\right)}$ because

$\displaystyle {f{{(}}}$$\displaystyle {)}\le$ $\displaystyle \le{f{{(}}}$$\displaystyle {)}$.