# Thread: Calculus & Analysis Marathon & Questions

1. ## Calculus & Analysis Marathon & Questions

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:

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

2. ## Re: First Year Uni Calculus Marathon

I can't word properly yet.

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

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

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

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

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

3. ## Re: First Year Uni Calculus Marathon

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

4. ## Re: First Year Uni Calculus Marathon

lel these look like tute or lecture qns

(I actually remember the above question haha)

5. ## Re: First Year Uni Calculus Marathon

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

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

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

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

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

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

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

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

6. ## Re: First Year Uni Calculus Marathon

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

7. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
Is this thread's main intention as a Q&A, or as a challenge thread?
I think exercise/challenge

My cries for help are going to remain in seperate threads

8. ## Re: First Year Uni Calculus Marathon

(A bit tricky)

For which positive integers k is it possible to find a continuous function f:R->R that such that

f(x)=y

has exactly k solutions x for every real number y?

9. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
(A bit tricky)

For which positive integers k is it possible to find a continuous function f:R->R that such that

f(x)=y

has exactly k solutions x for every real number y?
Is there any answer besides k=1?

10. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Is there any answer besides k=1?
Indeed that is the essence of the question, I don't want to spoil it too quickly. It is very good exercise to reach a conclusion yourself and try to rigorously justify it. (The main tools at your disposal being properties of continuous functions like the intermediate and extreme value theorem).

11. ## Re: First Year Uni Calculus Marathon

Originally Posted by Trebla
lel these look like tute or lecture qns

(I actually remember the above question haha)
There was a tute question similar to that question, with x^2 instead of x^3. (But it's the same general process.)

Originally Posted by InteGrand
Is this thread's main intention as a Q&A, or as a challenge thread?
Challenge/exercise, just like the other maths marathons

12. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Is there any answer besides k=1?
But what about k = ∞ ?

13. ## Re: First Year Uni Calculus Marathon

$k \in \mathbb Z^+$

$\Rightarrow f must be surjective (not necessarily useful).$

$Suppose k=2\\Then f(x)=y has two solutions for x. Suppose the values are at a, b.\\ Then, by Rolle's theorem: \exists c \in [a,b] \Rightarrow f^\prime (c)=0$

$If the entire interval was horizontal, then k immediately becomes undefined. So we can disregard this unique case.$

$Case 1: Where f(c)=d there exists no other x besides x=c such that f(x)=d\\This is a contradiction, as then we would require k=1. Thus this case cannot hold.$

$Case 2: Where f(c)=d there exists at least one other x besides x=c such that f(x)=d.\\ wlog. assume that such value for x (say, z) satisfies z>c\\Then, applying Rolle's theorem again\\ \exists h \in [c,z] \Rightarrow f^\prime(z)=0$

$However, let \delta represent the infinitesimal value.\\ In case 1, it was quite safe to assume that d-\delta (or d+\delta if c was the point of a local minimum) has exactly 2 solutions for f(x)=d-\delta$
$In case 2, however, since f(z)=d, there must also exist a third value such that f(x)=d-\delta (or alternatively d+\delta). Hence, this case also yields a contradiction as then we must have k=3$

$The proof can be replicated for all k>2. Thus, the only solution is k=1$

I hope...

14. ## Re: First Year Uni Calculus Marathon

But what about k = ∞ ?
Infinity is a concept not a number right

15. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
$k \in \mathbb Z^+$

$\Rightarrow f must be surjective (not necessarily useful).$

$Suppose k=2\\Then f(x)=y has two solutions for x. Suppose the values are at a, b.\\ Then, by Rolle's theorem: \exists c \in [a,b] \Rightarrow f^\prime (c)=0$

$If the entire interval was horizontal, then k immediately becomes undefined. So we can disregard this unique case.$

$Case 1: Where f(c)=d there exists no other x besides x=c such that f(x)=d\\This is a contradiction, as then we would require k=1. Thus this case cannot hold.$

$Case 2: Where f(c)=d there exists at least one other x besides x=c such that f(x)=d.\\ wlog. assume that such value for x (say, z) satisfies z>c\\Then, applying Rolle's theorem again\\ \exists h \in [c,z] \Rightarrow f^\prime(z)=0$

$However, let \delta represent the infinitesimal value.\\ In case 1, it was quite safe to assume that d-\delta (or d+\delta if c was the point of a local minimum) has exactly 2 solutions for f(x)=d-\delta$
$In case 2, however, since f(z)=d, there must also exist a third value such that f(x)=d-\delta (or alternatively d+\delta). Hence, this case also yields a contradiction as then we must have k=3$

$The proof can be replicated for all k>2. Thus, the only solution is k=1$

I hope...
For one thing, what if f need not be differentiable?

16. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
$We first request that f is continuous at x=0$

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

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

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

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

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

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

$Add on question: Show that the relevant definition is satisfied.$
Part 1 (not a very rigorous proof, though):

$As x\rightarrow \infty , x^3 + 1 \rightarrow x^3$
$As x\rightarrow \infty , x^2 + 1 \rightarrow x^2$
$Thus, \lim_{x \rightarrow \infty} \frac{\ln (x^3 + 1)}{\ln (x^2 + 1)} = \frac{\ln(x^3)}{\ln(x^2)} = \frac{3\ln(x)}{2\ln(x)} = \frac{3}{2}$

17. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
For one thing, what if f need not be differentiable?
Hmm...

$Cases where differentiability arise out of a discontinuity can be safely disregarded as the question already specified f is continuous on \mathbb R$

$This means that the lack of differentiability arises in the event of a cusp or a corner. However, that same cusp/corner will essentially also be a local extrema, thus the proof proceeds from there in a similar manner.$

At least, I don't think a cusp/corner can't be a local extrema here?

18. ## Re: First Year Uni Calculus Marathon

But what about k = ∞ ?
∞ is not a number, so k cannot equal ∞ right?

19. ## Re: First Year Uni Calculus Marathon

$To verify that k=1 works, any function that is bijective suffices$

For completeness sake

20. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Infinity is a concept not a number right
Originally Posted by Flop21
∞ is not a number, so k cannot equal ∞ right?
An equation can still have infinite solutions.

21. ## Re: First Year Uni Calculus Marathon

An equation can still have infinite solutions.
For this question, because infinity is not a measurable quantity (and I'm not sure if you can assume that it is an integer) we can't say for sure what the behaviour of f is if f(x)=y yields an infinite number of solutions.

At least, my simple brain can't visualise f anymore

22. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
$We first request that f is continuous at x=0$

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

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

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

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

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

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

$Add on question: Show that the relevant definition is satisfied.$
Also, regarding your proof, I think you need to the read the question again. (Particularly the definition of f(x))

23. ## Re: First Year Uni Calculus Marathon

Not even sure what I was thinking at the time now

24. ## Re: First Year Uni Calculus Marathon

Originally Posted by porcupinetree
Part 1 (not a very rigorous proof, though):

$As x\rightarrow \infty , x^3 + 1 \rightarrow x^3$
$As x\rightarrow \infty , x^2 + 1 \rightarrow x^2$
$Thus, \lim_{x \rightarrow \infty} \frac{\ln (x^3 + 1)}{\ln (x^2 + 1)} = \frac{\ln(x^3)}{\ln(x^2)} = \frac{3\ln(x)}{2\ln(x)} = \frac{3}{2}$
Here's my attempt:

Since x is positive, we can invoke the following inequalities:

$\noindent x^2 < x^2+1 < x^2 + 2x + 1 \\ x^3 < x^3 + 1 < x^3 + 3x^2 + 3x +1$

$\noindent After some muddling around, we get the following: \\ \frac{3\log{x}}{2\log{(x+1)}} < L < \frac{3\log{(x+1)}}{2\log{x}}$

It is easy to show that the left hand side is bounded from above by 3/2, and the right hand side is bounded from below by the same value.

It is then sufficient to show the left bound is monotonically increasing, and the right bound is monotonically decreasing, for sufficiently large x.

25. ## Re: First Year Uni Calculus Marathon

Here's my attempt:

Since x is positive, we can invoke the following inequalities:

$\noindent x^2 < x^2+1 < x^2 + 2x + 1 \\ x^3 < x^3 + 1 < x^3 + 3x^2 + 3x +1$

$\noindent After some muddling around, we get the following: \\ \frac{3\log{x}}{2\log{(x+1)}} < L < \frac{3\log{(x+1)}}{2\log{x}}$

It is easy to show that the left hand side is bounded from above by 3/2, and the right hand side is bounded from below by the same value.

It is then sufficient to show the left bound is monotonically increasing, and the right bound is monotonically decreasing, for sufficiently large x.
Believe it or not my workbook was happy if you just used L'Hopitals

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