Let's do some calculus! (52)

Calculus Level 4

Let $f(x)$ be a real-valued function which is positive and continuous in $[a,b]$ and differentiable in $(a,b)$ (where $b>a$ ). Given that

$\begin{cases} \displaystyle \lim_{x \to a^{+}} f(x) = 1 \\ \displaystyle \lim_{x \to b^{-}} f(x) = \sqrt[4] {3} \end{cases}$

and $f'(x) \ge {(f(x))}^3 + \dfrac{1}{f(x)}$ , find $(b-a)_{\text{max}}$ .

For more problems on calculus, click here .

\dfrac{\pi}{12}

\dfrac{\pi}{96}

\dfrac{\pi}{36}

\dfrac{\pi}{24}

1 solution

Tom Engelsman
Aug 28, 2017

Let $y = f(x)$ such that $\frac{dy}{dx} \ge y^3 + \frac{1}{y}$ , which separates into $\frac{y}{y^4 + 1} dy \ge dx$ . On the LHS, let $u = y^2, du = 2y dy$ so that we now have:

$\frac{1}{2} \cdot \frac{1}{u^2 + 1} du \ge dx \Rightarrow \frac{1}{2} \cdot arctan(u) \ge x + C \Rightarrow \boxed{\frac{1}{2} \cdot arctan(y^2) \ge x + C}.$

We now apply the original limits of $f(a) = 1, f(b) = 3^{\frac{1}{4}}$ which now yield:

$\frac{1}{2} \cdot arctan(1) \ge a + C \Rightarrow \frac{\pi}{8} \ge a + C;$

$\frac{1}{2} \cdot arctan (\sqrt{3}) \ge b + C \Rightarrow \frac{\pi}{6} \ge b + C.$

Finally, $(b-a)_{max} = (\frac{\pi}{6} - C) - (\frac{\pi}{8} - C) = \frac{4\pi - 3\pi}{24} = \boxed{\frac{\pi}{24}}.$

Exactly the same!!!

Aaghaz Mahajan - 3 years, 3 months ago

Good to know, Aaghaz, thanks :)

tom engelsman - 3 years, 3 months ago

Let's do some calculus! (52)

For more problems on calculus, click here .

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