Theoretical Recall

A function $f$ is differentiable at $x_0$ if the limit

\[f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}\]

exists and is finite. Differentiability implies continuity, but not vice versa.

  • Corners/cusps: arise when one-sided derivatives differ or blow up.
  • Vertical tangents: derivative tends to $\pm\infty$.
  • Piecewise definitions: for differentiability at the junction, require both continuity and equality of left/right derivatives.
  • Higher smoothness: $C^k(\mathbb{R})$ denotes $k$-times continuously differentiable functions.

Exercises

Exercise 1

Verify continuity and differentiability of

\[f(x)= \begin{cases} x^2, & x\ge 0, \\ -x, & x<0. \end{cases}\]

Solution:
At $0$, $\lim_{x\to0^+}f(x)=0=\lim_{x\to0^-}f(x)=f(0)$.
Left derivative: $\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{-h}{h}=-1$.
Right derivative: $\lim_{h\to0^+}\frac{h^2}{h}=0$.
Since they differ, $f$ is continuous but not differentiable at $0$.

Final Result:

\[f \in C^0(\mathbb{R}), \quad f \notin C^1(\mathbb{R}) \text{ (not differentiable at } 0).\]

Exercise 2

Study differentiability of $f(x)=x\sqrt{|x|}$.

Solution:
For $x>0$: $f(x)=x^{3/2}$, $f’(x)=\tfrac{3}{2}\sqrt{x}$.
For $x<0$: $f(x)=-x\sqrt{-x}=-(-x)^{3/2}$, $f’(x)=-\tfrac{3}{2}\sqrt{-x}$.
At $0$: $\lim_{h\to0^+}\frac{h^{3/2}}{h}=\sqrt{h}\to0$,
$\lim_{h\to0^-}\frac{-(-h)^{3/2}}{h}=\lim_{h\to0^+}\frac{h^{3/2}}{-h}=-\sqrt{h}\to0$.
So $f’(0)=0$.

Final Result:

\[f \in C^1(\mathbb{R}).\]

Exercise 3

Consider $f(x)=|x|\sqrt{|x|}$. Study differentiability at $0$.

Solution:
For $x>0$: $f(x)=x^{3/2}$, $f’(x)=\tfrac{3}{2}\sqrt{x}\to0$.
For $x<0$: $f(x)=(-x)^{3/2}$, derivative $f’(x)=-\tfrac{3}{2}\sqrt{-x}\to0$.
At $0$, both one-sided derivatives $=0$. Hence differentiable at $0$ with $f’(0)=0$.

Author’s note: Although the function involves absolute values and radicals, the exponents $>1$ guarantee smoothness at the origin.

Final Result:

\[f \in C^1(\mathbb{R}).\]

Exercise 4

Study differentiability of $f(x)=|x|\sqrt{1-x}$ at $x=1$.

Solution:
For $x<1$, $f$ is defined and continuous. At $x=1$: $f(1)=0$.
Right of 1 not in domain. Left derivative:
\(\lim_{h\to0^-}\frac{f(1+h)-0}{h}=\lim_{h\to0^-}\frac{|1+h|\sqrt{-h}}{h}.\) For $h\to0^-$, numerator $\sim \sqrt{-h}$, denominator $h<0$. Ratio $\to -\infty$.

Final Result:

\[f \text{ not differentiable at } 1 \text{ (vertical tangent)}.\]

Exercise 5

$f(x)=x^x\log x$ for $x>0$, extended by $f(0)=0$. Check differentiability at $0$.

Solution:
For $x>0$, write $f(x)=e^{x\log x}\log x$. Expansion: $x\log x\to0^-$ as $x\to0^+$, so $x^x\to1$. Then $f(x)\sim \log x$.
Thus $\lim_{x\to0^+}\frac{f(x)-0}{x}=\lim_{x\to0^+}\frac{\log x}{x}=-\infty$.

Final Result:

\[f \text{ is continuous at }0 \text{ but not differentiable (derivative } -\infty).\]

Exercise 6

$f(x)= \begin{cases} ax+b,& x<0,
x^2,& x\ge0. \end{cases}$
Find $a,b$ such that $f$ is differentiable at $0$.

Solution:
Continuity: $\lim_{x\to0^-}(ax+b)=b=f(0)=0 \Rightarrow b=0$.
Left derivative: $\lim_{h\to0^-}\frac{ah}{h}=a$.
Right derivative: $\lim_{h\to0^+}\frac{h^2}{h}=0$.
Equality $\Rightarrow a=0$.

Final Result:

\[a=0,\; b=0 \quad\Rightarrow f \in C^1(\mathbb{R}).\]

Exercise 7

For which $a\in\mathbb{R}$ is \(f(x)=|x|^a\) of class $C^k(\mathbb{R})$?

Solution:
Near $0$:

  • If $a>k$, derivatives up to order $k$ exist and are continuous.
  • If $a$ not integer, differentiability breaks at some finite order.

Author’s note: The function smoothness depends directly on the exponent.

Final Result:

\[f \in C^k(\mathbb{R}) \iff a>k.\]

Exercise 8

Find $a,b$ for which \(f(x)= \begin{cases} ax^2+bx,& x\ge0, \\ \sin x,& x<0 \end{cases}\) is continuous and differentiable at $0$.

Solution:
Continuity at 0: $\lim_{x\to0^-}\sin x=0=f(0)\Rightarrow$ automatic.
Derivative: left derivative $\cos 0=1$. Right derivative $\lim_{h\to0^+}\frac{ah^2+bh}{h}=b$.
So require $b=1$. $a$ free.

Final Result:

\[b=1,\; a \text{ arbitrary}.\]

Exercise 9

Find $a,b$ so that \(f(x)= \begin{cases} ax+b,& x<0, \\ x^2,& x\ge0 \end{cases}\) is $C^1$ at $0$.

Solution:
Continuity: $b=0$.
Derivative: left derivative $=a$, right derivative $=0$. Require $a=0$.

Final Result:

\[a=0,\; b=0.\]

Exercise 10

Find $a,b,c$ so that \(f(x)= \begin{cases} ax^2+bx+c,& x<0, \\ \cos x,& x\ge0 \end{cases}\) is $C^2(\mathbb{R})$.

Solution:
At $0$: $c=f(0)=\cos0=1$.
Derivative: left $f’(0^-)=b$, right $f’(0^+)= -\sin 0=0$. So $b=0$.
Second derivative: left $f’‘(0^-)=2a$, right $f’‘(0^+)= -\cos 0=-1$.
So $2a=-1 \Rightarrow a=-\tfrac{1}{2}$.

Final Result:

\[a=-\tfrac{1}{2},\; b=0,\; c=1.\]

Other Topics in Calculus I

These pages combine theoretical background and step-by-step worked examples.