Calculus & Analysis Marathon & Questions (2 Viewers)

porcupinetree · Apr 24, 2016

Calculus & Analysis Marathon & Questions
This is a marathon thread for single-variable calculus & analysis (mainly real and maybe complex analysis). Please aim to pitch your questions for first-year/second-year university level maths. Excelling & gifted/talented secondary school students are also invited to contribute.

(mod edit 7/6/17 by dan964)
===============================

Thought it'd be a good idea to start a marathon for users to post and answer first year uni calculus problems.

First question:

$Use Rolle's Theorem and the Intermediate Value Theorem to show that the equation $ \ln x + \frac{1}{x} = 2 $ has exactly 2 real solutions.$

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

I can't word properly yet.

$Consider the function $f : (0 , \infty) \rightarrow \mathbb R, f(x)=\ln{x}+\frac{1}{x}-2\\$Note that $f$ is continuous and differentiable everywhere.$

$f^{\prime}(x)=\frac{1}{x}-\frac{1}{x^2}\\$Find when $f^{\prime}$ vanishes to determine any local extrema on the curve:$\\ \frac{1}{x}-\frac{1}{x^2}=0 \Leftrightarrow x=1$

$Rolle's theorem states that if $f$ is continuous on [a,b] and differentiable on (a,b), and that $f(a)=f(b)$ then there exists at least one local extrema within that interval. Applying this, $f$ has at most two zeroes, since it has only one local extrema.$

$Given the point where the stationary point exists, consider the (continuous) intervals $(0,1] $ and $[1,\infty)\\$As $x\rightarrow 0, f(x) \rightarrow \infty\\f(1)=-1\\$Hence, by the intermediate value theorem there exists one solution to $f(x)=0$ in this interval (\because -1 < 0 < \infty).\\As $x\rightarrow \infty, f(x) \rightarrow \infty\\$ Hence, the IVT also anticipates one solution in this interval as well.$

$Thus $f(x)=0 \Leftrightarrow \ln{x}+\frac{1}{x}=2$ has exactly 2 solutions.$

porcupinetree · Apr 24, 2016

Re: First Year Uni Calculus Marathon

$Next question:$
$Show that $f$, where $f$ is defined by$
$f(x) = \left\{\begin{matrix} x^3 & $if $x$ is rational$ \\ 0 & $if $x$ is irrational$ \end{matrix}\right.$
$is differentiable at $x=0$.$

Trebla · Apr 24, 2016

Re: First Year Uni Calculus Marathon

lel these look like tute or lecture qns

(I actually remember the above question haha)

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

$$We first request that $f$ is continuous at $x=0$

$\\ \lim_{x\rightarrow 0^+}f(x) = \lim_{x\rightarrow 0^+}{x}^{3} =0\\\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^=}{x}^{3} =0 \\ \therefore \lim_{x\rightarrow 0}f(x) =0$

$\\ $So since $\lim_{x\rightarrow 0}f(x) = f(0) = 0\\ f(x)$ is indeed continuous at 0$$

$The derivative of a piecewise function cannot be computed by immediately taking derivatives at the point where $f$ changes it's rule.$

$\\ \lim _{ h\rightarrow { 0 }^{ + } }{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } } =\lim _{ h\rightarrow { 0 }^{ + } }{ \frac { { h }^{ 3 }-{ 0 }^{ 3 } }{ h } } =\lim _{ h\rightarrow { 0 }^{ + } }{ { h }^{ 2 } } =0\\ \lim _{ h\rightarrow { 0 }^{ - } }{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } } =\lim _{ h\rightarrow { 0 }^{ - } }{ \frac { { h }^{ 3 }-{ 0 }^{ 3 } }{ h } } =\lim _{ h\rightarrow { 0 }^{ - } }{ { h }^{ 2 } } =0\\ \therefore \lim _{ h\rightarrow { 0 } }{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } } =f^{ { \prime } }\left( 0 \right) =0$

$Hence, as the slope of the tangent is similar from both sides, $f$ must also be differentiable at 0 (or in other words, $f^\prime (0)$ is defined)$
=======================
$Next question$

$$Determine the behaviour of $f(x)=\frac{\ln{\left(x^3+1\right)}}{\ln{\left(x^2+1\right)}}$ as $x\rightarrow \infty$

$Add on question: Show that the relevant definition is satisfied.$

InteGrand · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Is this thread's main intention as a Q&A, or as a challenge thread?

leehuan · Apr 24, 2016