Question

worksheet 5: section 3.1 and 3.2: the derivative and differentiable functions
instructions: work together in a team of 3 or 4 to complete the problems. each team member should write their own solutions, neatly on separate sheets of paper, to be handed in at the end of discussion.
note: only handwritten work on paper in pencil or pen is accepted; electronic submissions (tablet, laptop, etc.) are not permitted.
note: even if you have had calculus before, we are assessing your skill to determine the limits algebraically. using lhopitals rule will not be given any credit.

1. use the limit definition of a derivative, \\(\lim_{t\to x}\frac{f(t)-f(x)}{t - x}\\), to find \\(f(-2)\\) for \\(f(x)=\frac{1}{x}\\).
2. let \\(g(x)=\frac{1}{x - 1}\\). use the definition of the derivative to evaluate \\(g(5)\\).
3. use the limit definition of a derivative to find \\(k(x)\\) for \\(k(x)=\frac{1}{x^{2}}-\sqrt{7}\\).
4. let \\(f(x)=\frac{3x + 1}{x + 5}\\). find \\(f(x)\\) using the limit definition of the derivative.
5. let \\(f(x)=\sqrt{x + 3}\\).

(a) use the limit definition of the derivative to find \\(f(x)\\).
(b) find the equation of the tangent line to \\(y = f(x)\\) at \\(x = 6\\).

1. below is the graph of a function \\(f\\). determine all values of \\(x\\) at which \\(f\\) is not differentiable. provide an explanation.

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{t
ightarrow x}\frac{f(t)-f(x)}{t - x}$.

Problem 1

Given $f(x)=\frac{1}{x}$, we want to find $f^{\prime}(-2)$.
\[

$$\begin{align*} f^{\prime}(x)&=\lim_{t ightarrow x}\frac{f(t)-f(x)}{t - x}=\lim_{t ightarrow x}\frac{\frac{1}{t}-\frac{1}{x}}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{x - t}{tx}}{t - x}=\lim_{t ightarrow x}\frac{-(t - x)}{tx(t - x)}\\ &=-\frac{1}{x^{2}} \end{align*}$$

\]
Substitute $x=-2$ into $f^{\prime}(x)$, we get $f^{\prime}(-2)=-\frac{1}{4}$.

Problem 2

Given $g(x)=\frac{1}{x - 1}$, then
\[

$$\begin{align*} g^{\prime}(x)&=\lim_{t ightarrow x}\frac{g(t)-g(x)}{t - x}=\lim_{t ightarrow x}\frac{\frac{1}{t - 1}-\frac{1}{x - 1}}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{(x - 1)-(t - 1)}{(t - 1)(x - 1)}}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{x - t}{(t - 1)(x - 1)}}{t - x}\\ &=\lim_{t ightarrow x}\frac{-(t - x)}{(t - 1)(x - 1)(t - x)}\\ &=-\frac{1}{(x - 1)^{2}} \end{align*}$$

\]
Substitute $x = 5$ into $g^{\prime}(x)$, we get $g^{\prime}(5)=-\frac{1}{16}$.

Problem 3

Given $k(x)=\frac{1}{x^{2}}-\sqrt{7}$, then
\[

$$\begin{align*} k^{\prime}(x)&=\lim_{t ightarrow x}\frac{k(t)-k(x)}{t - x}=\lim_{t ightarrow x}\frac{(\frac{1}{t^{2}}-\sqrt{7})-(\frac{1}{x^{2}}-\sqrt{7})}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{1}{t^{2}}-\frac{1}{x^{2}}}{t - x}=\lim_{t ightarrow x}\frac{\frac{x^{2}-t^{2}}{t^{2}x^{2}}}{t - x}\\ &=\lim_{t ightarrow x}\frac{-(t - x)(t + x)}{t^{2}x^{2}(t - x)}\\ &=-\frac{2x}{x^{4}}=-\frac{2}{x^{3}} \end{align*}$$

\]

Problem 4

Given $f(x)=\frac{3x + 1}{x + 5}$, then
\[

$$\begin{align*} f^{\prime}(x)&=\lim_{t ightarrow x}\frac{f(t)-f(x)}{t - x}=\lim_{t ightarrow x}\frac{\frac{3t+1}{t + 5}-\frac{3x + 1}{x + 5}}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{(3t + 1)(x + 5)-(3x + 1)(t + 5)}{(t + 5)(x + 5)}}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{3tx+15t+x + 5-(3tx+15x+t + 5)}{(t + 5)(x + 5)}}{t - x}\\ &=\lim_{t ightarrow x}\frac{\frac{14(t - x)}{(t + 5)(x + 5)}}{t - x}\\ &=\frac{14}{(x + 5)^{2}} \end{align*}$$

\]

Problem 5

(a) Given $f(x)=\sqrt{x + 3}$, then
\[

$$\begin{align*} f^{\prime}(x)&=\lim_{t ightarrow x}\frac{f(t)-f(x)}{t - x}=\lim_{t ightarrow x}\frac{\sqrt{t + 3}-\sqrt{x + 3}}{t - x}\\ &=\lim_{t ightarrow x}\frac{(\sqrt{t + 3}-\sqrt{x + 3})(\sqrt{t + 3}+\sqrt{x + 3})}{(t - x)(\sqrt{t + 3}+\sqrt{x + 3})}\\ &=\lim_{t ightarrow x}\frac{(t + 3)-(x + 3)}{(t - x)(\sqrt{t + 3}+\sqrt{x + 3})}\\ &=\lim_{t ightarrow x}\frac{t - x}{(t - x)(\sqrt{t + 3}+\sqrt{x + 3})}\\ &=\frac{1}{2\sqrt{x+3}} \end{align*}$$

\]
(b) First, when $x = 6$, $f(6)=\sqrt{6 + 3}=3$. And $f^{\prime}(6)=\frac{1}{2\sqrt{6 + 3}}=\frac{1}{6}$.
The equation of the tangent - line is $y - y_{0}=m(x - x_{0})$, where $(x_{0},y_{0})=(6,3)$ and $m = f^{\prime}(6)=\frac{1}{6}$.
So the equation of the tangent - line is $y-3=\frac{1}{6}(x - 6)$, which simplifies to $y=\frac{1}{6}x+2$.

Problem 6

A function $y = f(x)$ is not differentiable at a point $x=a$ if:

1. There is a sharp corner at $x=a$.
2. There is a vertical tangent at $x=a$.
3. The function is discontinuous at $x=a$.

Looking at the graph, the function $f(x)$ is not differentiable at $x=-3$ (sharp corner) and $x = 3$ (vertical tangent).

Answer:

1. $f^{\prime}(-2)=-\frac{1}{4}$
2. $g^{\prime}(5)=-\frac{1}{16}$
3. $k^{\prime}(x)=-\frac{2}{x^{3}}$
4. $f^{\prime}(x)=\frac{14}{(x + 5)^{2}}$
5. (a) $f^{\prime}(x)=\frac{1}{2\sqrt{x+3}}$; (b) $y=\frac{1}{6}x + 2$
6. The function $f$ is not differentiable at $x=-3$ (sharp corner) and $x = 3$ (vertical tangent).