A few Calculus Problems to get more ideas from

by SomeonecoolLovesMaths, Nov 29, 2024, 6:24 AM

$f$ is a continuous function on $[a,b]$ , differentiable on $(a,b)$ , and $f(a) = f(b) = 0$ . For any $\alpha \in \mathbb{R}$ , show that $\exists c \in (a,b)$ such that $f'(c) + \alpha f(c)=0$

$\underline{\text{A 'stupid' method}}$
$f'(x) + \alpha f(x) = 0$ "for all $x$ "
$\Longrightarrow \int \frac{f'(x)}{f(x)} \text{d}x = \int - \alpha \text{d} x$
$\Longrightarrow \ln f(x) = -\alpha x + c$
$\Longrightarrow f(x) = e^{- \alpha x} \times \text{constant}$
$\Longrightarrow e^{\alpha x f(x)} = \text{constant}$
Apply Rolle's Theorem on $g(x) = e^{\alpha x} f(x)$ .

S1 by tobiSALT

Let $g(x) = e^{\alpha x} f(x)$ . Since $f(x)$ is continuous on $[a,b]$ and $e^{\alpha x}$ is continuous for all $x \in \mathbb{R}$ , their product $g(x)$ is continuous on $[a,b]$ .
Since $f(x)$ is differentiable on $(a,b)$ and $e^{\alpha x}$ is differentiable for all $x \in \mathbb{R}$ , their product $g(x)$ is differentiable on $(a,b)$ .
Moreover, $g(a) = e^{\alpha a} f(a) = e^{\alpha a} \cdot 0 = 0$ and $g(b) = e^{\alpha b} f(b) = e^{\alpha b} \cdot 0 = 0$ .
Since $g(x)$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $g(a) = g(b) = 0$ , we can apply Rolle's Theorem.
Therefore, there exists some $c \in (a,b)$ such that $g'(c) = 0$ .

Now we compute $g'(x)$ . Using the product rule, we have
$g'(x) = (e^{\alpha x})' f(x) + e^{\alpha x} f'(x) = \alpha e^{\alpha x} f(x) + e^{\alpha x} f'(x) = e^{\alpha x} (\alpha f(x) + f'(x))$ Since $g'(c) = 0$ , we have
$e^{\alpha c} (\alpha f(c) + f'(c)) = 0$ Since $e^{\alpha c} \neq 0$ for any $c \in (a,b)$ , we must have
$\alpha f(c) + f'(c) = 0$ This is equivalent to
$f'(c) + \alpha f(c) = 0$ Thus, there exists $c \in (a,b)$ such that $f'(c) + \alpha f(c) = 0$ .

Well, the expression $f'(c) + \alpha f(c)$ looks like it could come from the derivative of a product. To make this explicit, we search for an "integrating factor." We want a function whose derivative is a multiple of itself. This gives the idea of using an exponential function.
Let $g(x) = e^{\alpha x} f(x)$ . Then $g(a) = e^{\alpha a} f(a) = 0$ and $g(b) = e^{\alpha b} f(b) = 0$ since $f(a) = f(b) = 0$ .
We have $g'(x) = \alpha e^{\alpha x} f(x) + e^{\alpha x} f'(x) = e^{\alpha x} (f'(x) + \alpha f(x))$ .
Since $g(a) = g(b) = 0$ , by Rolle's theorem, there exists $c \in (a,b)$ such that $g'(c) = 0$ .
Thus $e^{\alpha c} (f'(c) + \alpha f(c)) = 0$ . Since $e^{\alpha c} \neq 0$ , we must have $f'(c) + \alpha f(c) = 0$ .

S1 by Levieee

Assume the equation
$f'(c) + \alpha f(c) = 0$ holds for all $c \in [a, b]$ . The intuition is that we want to apply Rolle's theorem on a function $h(x)$ in such a way that it gives us the required equation. To do this, a better approach is to backtrack and find $h(x)$ based on the given equation.

Now assume the general form $f'(x) + \alpha f(x) = 0$ . Great! We can integrate this equation to solve for $f(x)$ , but directly integrating will give a new function, i.e., the antiderivative of $f(x)$ . Instead, we reformulate it into something familiar.

We know that the derivative of $\ln(f(x))$ is $\frac{f'(x)}{f(x)}$ . Hence, rewriting:
$\frac{f'(x)}{f(x)} = -\alpha.$ Now integrate both sides (recalling that integration is the reverse of differentiation):
$\int \frac{f'(x)}{f(x)} \, dx = \int -\alpha \, dx.$ This gives:
$\ln(f(x)) = -\alpha x + C,$ where $C$ is the constant of integration.

To solve for $f(x)$ , exponentiate both sides:
$e^{\ln(f(x))} = e^{-\alpha x + C}.$ Thus:
$f(x) = e^{-\alpha x + C}.$ Using the properties of exponents, this simplifies to:
$f(x) = e^C e^{-\alpha x}.$ Now, observe that $f(x)$ itself is not the function we need to apply Rolle's theorem because its derivative does not necessarily vanish. To find the desired function, note that multiplying $f(x)$ by $e^{\alpha x}$ eliminates the exponential decay term:
$f(x) \cdot e^{\alpha x} = e^C.$ This gives us a constant function, which satisfies the condition for Rolle's theorem. Therefore, the function:
$h(x) = f(x) \cdot e^{\alpha x} = e^C$ is the one we are looking for.

https://www.youtube.com/watch?v=DJsjZ5aYK_g
watch this , this might be helpful

Prove that the equation $(1-x) \cos x = \sin x$ has atleast one solution in $(0,1)$ .

Consider $f(x) = \cos x - x \cos x - \sin x$ . Observe $f(0) = 1 > 0$ and $f \left ( \frac{\pi}{4}) \right) = -\frac{\pi}{4} \cdot \frac{1}{ \sqrt{2}} < 0$ . Thus by IVT we are done.

If $f: [0,1] \longrightarrow [0,1]$ is a continuous function, then show that there exists $c \in [0,1]$ such that $f(c) = c^2$ .

Let $g(x) = f(x) - x^2$ . $g(0) = f(0) - 0 \geq 0$ . $g(1) = f(1) - 1 \leq 0$ . If any equality occurs then we are already done. If not then apply IVT on $g$ to get the result.

Suppose that $f,g: \mathbb{R} \longrightarrow \mathbb{R}$ are continuous and such that $f(a) < g(a)$ and $f(b) > g(b)$ . Show that there exists $c \in (a,b)$ such that $f(c) = g(c)$ .

Consider $h(x) = f(x) - g(x)$ . $h(b) > 0 > h(a)$ and $h$ is continuous. Hence by IVT we are done.

Let $f:[a,b] \longrightarrow \mathbb{R}$ be a continuous function with the property that for every $x \in [a,b]$ , there exists $y \in [a,b]$ such that $|f(y)| \leq \frac{1}{2} |f(x)|$ . Show that there exists $c \in [a,b]$ such that $f(c) = 0$ .

Let $f$ have the same sign all over the range or we will be done by IVT. Say $f\geq 0$ (or else consider $-f$ ). Thus $f(y) = \frac{1}{2} f(x)$ . Now as $f$ goes from a closed interval to reals and is continuous so $f$ must be bounded, thus having a global minima by EVT. Say $f(c)$ is the global minima, then there exists $d$ such that $f(d) \leq \frac{1}{2} f(c) \leq f(c)$ (equality iff $f(c) = 0$ ). Now such a $d$ cannot exist since $f(c)$ is the global minima, hence $d=c$ and $f(c) = 0$ .

Suppose that $f: [0,1] \longrightarrow [0,1]$ is a continuous function with $f(0) = 0$ and $f(1) = 1$ . Show that there exists $c \in (0,1)$ such that $c^2 + f(c)^2 = 1$ .

Let $g(x) = f(x) - \sqrt{1 - c^2}$ . Note that this is possible as $c^2 <1$ . Now $g(0) = -1$ and $g(1) = 1$ . Thus by IVT we are done.

Let $f:[a,b] \longrightarrow \mathbb{R}$ be a continuous function. Let $x_1,x_2, \cdots , x_n$ be any $n$ points in $(a,b)$ . Show that there exists $x_0 \in (a,b)$ such that

WOLOG assume $f(x_1) \leq f(x_2) \leq f(x_3) \leq \cdots \leq f(x_n)$ , then $f(x_1) \leq \frac{1}{n} \left(f(x_1) + f(x_2) + \cdots f(x_n) \right) \leq f(x_n)$ . Thus by IVT we are done.

Suppose that $f$ and $f'$ are continuous functions on $\mathbb{R}$ and that $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to\infty}f'(x)$ exist. Show that $\displaystyle\lim_{x\to\infty}f'(x) = 0.$

Consider $g(x) = e^x f(x)$ and $h(x) = e^x$ . Then $\lim_{x \longrightarrow \infty} f(x) = \lim_{x \longrightarrow \infty} \frac{g(x)}{h(x)} = \lim_{x \longrightarrow \infty} \frac{g'(x)}{h'(x)} = \lim_{x \longrightarrow \infty} \frac{e^x (f(x) + f'(x))}{e^x} =\lim_{x \longrightarrow \infty} (f(x)+f'(x))$ . Hence $\lim_{x \longrightarrow \infty} f'(x) = 0$ .

Q9 (IMC 2016)

Let $f : \left[ a, b\right]\rightarrow\mathbb{R}$ be continuous on $\left[ a, b\right]$ and differentiable on $\left( a, b\right)$ . Suppose that $f$ has infinitely many zeros, but there is no $x\in \left( a, b\right)$ with $f(x)=f'(x)=0$ .
(a) Prove that $f(a)f(b)=0$ .
(b) Give an example of such a function on $\left[ 0, 1\right]$ .

Let the sequence of the roots be $\{ x_n \}_{n \geq 1}$ .

Lemma: Every sequence of real numbers has a monotone subsequence.
Proof: Let $S = \{ n \mid x_m > x_n \; \forall \; m>n \}$ . If $S$ is infinite $\{n_1,n_2,n_3, \cdots\}$ , then the sequence $x_{n_1} < x_{n_2} < x_{n_3} < \cdots$ is a monotone subsequence.
If $S$ is finite, then $\exists \; n_1$ such that $n < n_1$ $\forall$ $n \in S$ . Since $n_1 \not \in S$ , then $\exists$ a $n^2 > n_1$ such that $x_{n_2} \leq x_{n_1}$ . As $n_2 \not \in S$ , $\exists$ $n_3$ such that $x_{n_3} \leq x_{n_2} \leq x_{n_1}$ and so on.

Lemma: A monotone bounded sequence of real numbers converges.
Proof: WOLOG, assume that $x_n$ and $x$ is the supremum of $x_n$ . Our claim is that $x_n \longrightarrow x$ .
For any $\epsilon >0$ , by the definition of $\text{sup} \; x_n$ there exists $x_n$ such that $|x_n - x| < \epsilon$ . Set $N=n$ . Then, by monotonicity $x_m > x_N$ $\forall$ $m>n$ . But as $x_m < x$ , this implies that $|x_m - x| < \epsilon$ $\forall$ $m>N$ , so $x_n \longrightarrow x$ $\blacksquare$ .

Consider a monotonic subsequence of $\{ x_n \}_{n \geq 1}$ say $\{ y_n \}_{n \geq 1}$ . Thus it converges to some point $c \in [a,b]$ .
Say $c \in (a,b)$ .
As $f$ is continuous of $[a,b]$ thus $f(c) = 0$ . Now using the definition of differentiability, $\boxed{f'(c) = \lim_{x \longrightarrow c} \frac{f(x) - f(c)}{x-c}}.$ Notice the fact that for all sequences $a_{n}$ in $[a,b]$ converging to $c$ we must have $\lim_{a_{n} \rightarrow c} \frac{f(a_{n})-f(c)}{a_{n}-c}$ . Setting $\{ a_n \}_{n \geq 1} = \{ y_n \}_{n \geq 1}$ , we get $f'(c) = 0$ , impossible as per the given condition. Thus $c \in \{a,b\}$ and hence either $f(a) = 0$ or $f(b) = 0$ .

Consider the following function $\boxed{f(x)=\begin{cases} x\sin\frac{1}{x} & \mathrm{if }\, 0<x\le 1\\ 0 & \mathrm{if}\, x=0 \end{cases}}$ It is ofcourse continuous at $(0,1]$ .

Sandwich Theorem: Suppose $f(x)$ is between $g(x)$ and $h(x)$ for all $x$ in a neighborhood of the point $S$ . If $g$ and $h$ approach some common limit $L$ as $x$ approaches $S$ , then $\lim_{x\to S}f(x)=L$ .
Proof: If $f(x)$ is between $g(x)$ and $h(x)$ for all $x$ in the neighborhood of $S$ , then either $g(x)\leq f(x) \leq h(x)$ or $h(x)\leq f(x)\leq g(x)$ for all $x$ in this neighborhood. The two cases are the same up to renaming our functions, so assume without loss of generality that $g(x)\leq f(x) \leq h(x)$ .

We must show that for all $\varepsilon >0$ there is some $\delta > 0$ for which $|x-S|<\delta$ implies $|f(x)-L|<\varepsilon$ .

Now since $\lim_{x\to S}g(x)=\lim_{x\to S}h(x)=L$ , there must exist $\delta_1,\delta_2>0$ such that

$|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon \textrm{ and } |x-S|<\delta_2 \Rightarrow |h(x)-L|<\varepsilon.$
Now let $\delta = \min\{\delta_1,\delta_2\}$ . If $|x-S|<\delta$ then

$-\varepsilon < g(x) - L \leq f(x) - L \leq h(x) - L < \varepsilon.$

So $|f(x)-L|<\varepsilon$ . Now by the definition of a limit we get $\lim_{x\to S}f(x)=L$ as desired.

Observe $-1 \leq \sin \left( \frac{1}{x} \right) \leq 1$ . Thus $-x \leq x\sin \left( \frac{1}{x} \right) \leq x$ . As $x$ approaches $0$ , $g(x) = -x$ and $h(x) = x$ tend to $0$ . Thus using the Sandwich Theorem, $x\sin \left( \frac{1}{x} \right)$ will also approach $0$ .
Observe LHD of $x\sin \left( \frac{1}{x} \right)$ is $\lim_{h \longrightarrow \infty} - \sin \left( \frac{1}{h} \right)$ while RHD of $x\sin \left( \frac{1}{x} \right)$ is $\lim_{h \longrightarrow \infty} \sin \left( \frac{1}{h} \right)$ which means it is not differentiable at $0$ .

This post has been edited 2 times. Last edited by SomeonecoolLovesMaths, Dec 5, 2024, 10:38 AM

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for Q7 the condition is missing

by quasar_lord, Mar 12, 2025, 6:22 PM

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