Example 3.5.1.
If \(f'(a^+)=3, \) then find
\begin{equation*}
\lim_{h \to 0}\frac{f(a+2h^2) - f(a)}{h^2}
\end{equation*}
Section 3.5 Differentiability
The function \(f(x)\) is continuous at a point \(x=a,\) if
\begin{equation}
\lim_{x\to a} f(x) = f(a)\tag{3.5.1}
\end{equation}
Which also means Left hand derivative is equal to right hand derivative.
\begin{equation}
LHL\vert_a = RHL\vert_a = f(a)\tag{3.5.2}
\end{equation}
The function \(f(x)\) is differentiable at \(x=a,\) if
\begin{equation*}
LHD\vert_{x=a} = RHD\vert_{x=a} =finite
\end{equation*}
\begin{equation}
or, \, f'(a^-) = f'(a^+) = finite \tag{3.5.3}
\end{equation}
\begin{equation}
or, \, f'(x^-) = f'(x^+) \tag{3.5.4}
\end{equation}
but if
\begin{equation}
f'(x^-) \neq f'(x^+) \tag{3.5.5}
\end{equation}
then \(f(x)\) is not differentiable. If
\begin{equation*}
LHL\vert_{x=a} = RHL\vert_{x=a} =finite
\end{equation*}
then limit exists and if
\begin{equation*}
LHL\vert_{a} = RHL\vert_{a} =f(a),
\end{equation*}
then \(f(x)\) is continuous.
\begin{align*}
RHD\vert_{x=a} \amp = f'(a^+) = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}, \quad \, h \gt 0\\
LHD\vert_{x=a} \amp =f'(a^-) = \lim_{h \to 0}\frac{f(a-h) - f(a)}{-h}, \quad \, h \gt 0
\end{align*}
Example 3.5.2.
At what numbers is the following function \(g\) differentiable?
\begin{equation*}
g(x) = \begin {cases} 2x & \text{ if } x\leq 0 \\
2x-x^2 & \text{ if } 0\lt x \lt 2 \\
2-x & \text{ if } x \geq 2
\end {cases}
\end{equation*}
Solution.
A function \(f\) is differentiable at \(a\) if the function is continuous, i.e., \(f(a^-) = f(a^+) = f(a)\)
At \(x=0\)
\begin{equation*}
\lim_{x \to 0} g(x) = g(0) = 0.
\end{equation*}
\begin{equation*}
also, \, \lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} 2x = 0.
\end{equation*}
\begin{equation*}
and, \, \lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} 2x-x^2 = 2\cdot 0 -0^2 =0.
\end{equation*}
Hence, the function is continuous at \(x=0.\)
At \(x=2\)
\begin{equation*}
\lim_{x \to 2} g(x) = \lim_{x \to 2} (2-x) = 2-2 =0.
\end{equation*}
\begin{equation*}
also, \, \lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} (2x-x^2) = 2\cdot 2-2^2= 0.
\end{equation*}
\begin{equation*}
and, \, \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (2-x) = 2-2 =0.
\end{equation*}
Hence, the function is continuous at \(x=2.\) Polynomials are differentiable everywhere, but need to check differentiabilty at the given point.
Now Left hand derivative at \(x=0.\)
\begin{equation*}
g(x) = 2x \, \Rightarrow g'(x)= 2 \, \Rightarrow g'(0)= 2.
\end{equation*}
again, Right hand derivative at \(x=0.\)
\begin{equation*}
g(x) = 2x-x^2 \, \Rightarrow g'(x)= 2-2x \, \Rightarrow g'(0)= 2.
\end{equation*}
Hence, the function is continuous and left derivative = right derivative = 2.
Now check differentiabilty at \(x=2.\)
\begin{equation*}
LHD\vert_{x=2}, \, g(x) = 2x-x^2 \, \Rightarrow g'(x)= 2-2x \, \Rightarrow g'(2^-)= 2-2\cdot 2= -2.
\end{equation*}
\begin{equation*}
RHD\vert_{x=2}, \, g(x) = 2-x \, \Rightarrow g'(x)= -1 \, \Rightarrow g'(2^+)= -1.
\end{equation*}
That is,
\begin{equation*}
g'(2^+) \neq g'(2^-)
\end{equation*}
Hence, the function is not differentiable at \(x=2, \) however, it is continous at \(x=2.\)
Example 3.5.3.
Let
\begin{equation*}
f(x) = \begin {cases} x^2+1 & \text{ if } x\lt 1 \\
x+1 & \text{ if } x \geq 1 \\
\end {cases}
\end{equation*}
is \(f\) differentiable at \(1?\) Sketch the graph of \(f\) and \(f'.\)
Solution.
Polynomials are differentiable everywhere, just need to check differentiability at \(x=1.\)
\begin{align*}
LHL\vert_{x=1}, \, f(x) \amp = x^2+1 \, \Rightarrow f(1^-)= 1+1 =2. \\
LHD\vert_{x=1},\, f'(x) \amp = 2x \, \Rightarrow f'(1^-)= 2\\cdot 1 = 2.
\end{align*}
\begin{align*}
RHL\vert_{x=1}, \, f(x) \amp = x+1 \, \Rightarrow f(1^+)= 1+1 =2. \\
RHD\vert_{x=1},\, f'(x) \amp = 1 \, \Rightarrow f'(1^+)= 1.
\end{align*}
Hence, the function is continuous as \(f(x)=f(x^-)= f(x^+)\) but not differentiable as \(f'(x^-)\neq f'(x^+)\text{.}\)
Example 3.5.4.
In
\begin{equation*}
f(x) = \begin {cases} x^2 & \text{ if } x \leq 2 \\
mx+b & \text{ if } x \gt 2 \\
\end {cases}
\end{equation*}
find \(m\) and \(b.\)
Solution.
\begin{align*}
LHL\vert_{x=2}, \, f(x) \amp = x^2 \, \Rightarrow f(2^-)= 2^2 =4. \\
LHD\vert_{x=2},\, f'(x) \amp = 2x \, \Rightarrow f'(2^-)= 2\\cdot 2 = 4.
\end{align*}
\begin{align*}
RHL\vert_{x=2}, \, f(x) \amp = mx+b \, \Rightarrow f(x^+)= m\cdot 2+b =2m+b. \\
RHD\vert_{x=2}, \, f'(x) \amp = m \, \Rightarrow f'(m^+)= m.
\end{align*}
Also \(f(x) = x^2\) so \(f(2)=2^2=4.\) i.e., the given function must be continous and differentiable in order to determine the value of \(m\) and \(b\text{.}\) Hence, to be differentiable
\begin{align*}
LHD\vert_{x=2} \amp = RHD\vert_{x=2} \\
m \amp =4
\end{align*}
and to be continuous
\begin{align*}
LHL\vert_{x=2} \amp = RHL\vert_{x=2} \\
2m+b \amp = 4 \, \Rightarrow \, b= 4-2m = 4-8=-4.
\end{align*}
