Practice questions

Here ${f(x)=x^2+3x}$ and ${a=1}$, so ${f(1)=4}$. Use the slope of the tangent line:

$\begin{align*}&m\\&=\lim_{h\to 0}\dfrac{f(1+h)-f(1)}{h}\\&=\lim_{h\to 0}\dfrac{(1+h)^2+3(1+h)-4}{h}\end{align*}$

Expand ${(1+h)^2}$ and ${3(1+h)}$. The numbers cancel:

$\begin{align*}&=\lim_{h\to 0}\dfrac{1+2h+h^2+3+3h-4}{h}\\&=\lim_{h\to 0}\dfrac{5h+h^2}{h}\end{align*}$

Cancel the common factor ${h}$:

$\begin{align*}&=\lim_{h\to 0}(5+h)\end{align*}$

Let ${h}$ approach ${0}$:

$\begin{align*}&m=5\end{align*}$

The tangent line goes through ${(1,4)}$ with slope ${5}$:

$\begin{align*}y-4&=5(x-1)\\y&=5x-1\end{align*}$

Here ${f(x)=\dfrac{2}{x}}$ and ${a=1}$, so ${f(1)=2}$. Use the slope of the tangent line:

$\begin{align*}&m\\&=\lim_{h\to 0}\dfrac{\dfrac{2}{1+h}-2}{h}\end{align*}$

Combine the fractions in the numerator, using the common denominator ${1+h}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{\dfrac{2-2(1+h)}{1+h}}{h}\end{align*}$

Simplify the numerator. Dividing by ${h}$ puts ${h}$ in the denominator:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-2h}{h(1+h)}\end{align*}$

Cancel the common factor ${h}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-2}{1+h}\end{align*}$

Let ${h}$ approach ${0}$:

$\begin{align*}&m=-2\end{align*}$

The tangent line goes through ${(1,2)}$ with slope ${-2}$:

$\begin{align*}y-2&=-2(x-1)\\y&=-2x+4\end{align*}$

Find the heights at the two times:

${s(1)=20(1)-5(1)^2=15}$

${s(2)=20(2)-5(2)^2=20}$

The average velocity is the change in position divided by the change in time:

$\begin{align*}&\text{average velocity}\\&=\dfrac{s(2)-s(1)}{2-1}\\&=\dfrac{20-15}{1}\\&=5\ \text{m/s}\end{align*}$

The velocity at ${t=1}$ is the limit of the average velocity:

$\begin{align*}&v(1)\\&=\lim_{h\to 0}\dfrac{s(1+h)-s(1)}{h}\end{align*}$

First, work out ${s(1+h)}$:

$\begin{align*}&s(1+h)\\&=20(1+h)-5(1+h)^2\\&=20+20h-5(1+2h+h^2)\\&=20+20h-5-10h-5h^2\\&=15+10h-5h^2\end{align*}$

Subtract ${s(1)=15}$. The ${15}$s cancel:

$\begin{align*}&v(1)\\&=\lim_{h\to 0}\dfrac{10h-5h^2}{h}\end{align*}$

Cancel the common factor ${h}$, and let ${h}$ approach ${0}$:

$\begin{align*}&=\lim_{h\to 0}(10-5h)\\&=10\ \text{m/s}\end{align*}$

First, ${s(3)=20(3)-5(3)^2=15}$. Next, work out ${s(3+h)}$:

$\begin{align*}&s(3+h)\\&=20(3+h)-5(3+h)^2\\&=60+20h-5(9+6h+h^2)\\&=60+20h-45-30h-5h^2\\&=15-10h-5h^2\end{align*}$

Subtract ${s(3)=15}$. The ${15}$s cancel:

$\begin{align*}&v(3)\\&=\lim_{h\to 0}\dfrac{-10h-5h^2}{h}\end{align*}$

Cancel the common factor ${h}$, and let ${h}$ approach ${0}$:

$\begin{align*}&=\lim_{h\to 0}(-10-5h)\\&=-10\ \text{m/s}\end{align*}$

The velocity is negative, so at ${t=3}$ the ball is moving down. It moves at a speed of ${10}$ m/s.

Step 1: ${f(2)=3(2)^2-2=10}$.

Step 2: Expand ${f(2+h)}$:

$\begin{align*}&f(2+h)\\&=3(2+h)^2-(2+h)\\&=3(4+4h+h^2)-2-h\\&=12+12h+3h^2-2-h\\&=10+11h+3h^2\end{align*}$

Step 3: Find the numerator of the difference quotient:

$\begin{align*}&f(2+h)-f(2)\\&=(10+11h+3h^2)-10\\&=11h+3h^2\end{align*}$

Step 4: Divide by ${h}$, and let ${h}$ approach ${0}$:

$\begin{align*}&f'(2)\\&=\lim_{h\to 0}\dfrac{11h+3h^2}{h}\\&=\lim_{h\to 0}(11+3h)\\&=11\end{align*}$

The tangent line goes through ${(2,10)}$ with slope ${11}$:

$\begin{align*}y-10&=11(x-2)\\y-10&=11x-22\\y&=11x-12\end{align*}$

Here ${f(1)=\dfrac{1}{3}}$ and $f(1+h)=\dfrac{1}{2(1+h)+1}=\dfrac{1}{3+2h}$. Use the definition of the derivative:

$\begin{align*}&f'(1)\\&=\lim_{h\to 0}\dfrac{\dfrac{1}{3+2h}-\dfrac{1}{3}}{h}\end{align*}$

Combine the two fractions in the numerator, using the common denominator ${3(3+2h)}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{\dfrac{3-(3+2h)}{3(3+2h)}}{h}\end{align*}$

Simplify the numerator. Dividing by ${h}$ puts ${h}$ in the denominator:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-2h}{3h(3+2h)}\end{align*}$

Cancel the common factor ${h}$, and let ${h}$ approach ${0}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-2}{3(3+2h)}\\&=\dfrac{-2}{3(3)}\\&=-\dfrac{2}{9}\end{align*}$

Here ${f(5)=25}$. Use the ${x\to a}$ form of the derivative:

$\begin{align*}&f'(5)\\&=\lim_{x\to 5}\dfrac{x^2-25}{x-5}\end{align*}$

The numerator is a difference of squares. Factor it:

$\begin{align*}&=\lim_{x\to 5}\dfrac{(x-5)(x+5)}{x-5}\end{align*}$

Cancel the common factor ${x-5}$, and let ${x}$ approach ${5}$:

$\begin{align*}&=\lim_{x\to 5}(x+5)\\&=10\end{align*}$

${f'(1)}$ is the slope of the tangent line. The tangent line goes through ${(1,3)}$ and ${(2,1)}$. From ${(1,3)}$ to ${(2,1)}$, the run is ${1}$ and the rise is ${-2}$ (the line goes down). So

$f'(1)=\dfrac{\text{rise}}{\text{run}}=\dfrac{-2}{1}=-2$

At ${x=-1}$, the graph of ${f}$ is rising (going up from left to right). So the slope there is positive, and ${f'(-1)>0}$.

The tangent line is horizontal at the top of the graph, where ${x=0}$. A horizontal line has slope ${0}$. So ${f'(0)=0}$.

${T'(5)}$ is the rate of change of the temperature at ${t=5}$ minutes. Its units are degrees Fahrenheit per minute. The negative sign means the temperature is going down.

So, ${5}$ minutes after the coffee is poured, it is cooling at a rate of ${3\,^\circ}$F per minute.

${C'(100)}$ is the rate of change of the cost when ${x=100}$. Its units are dollars per T-shirt.

So, after ${100}$ T-shirts are made, the cost is going up at a rate of ${8}$ dollars per T-shirt. Making one more T-shirt costs about ${8}$ dollars.

Use the definition of the derivative:

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\end{align*}$

Work out the numerator. The ${x^2}$ terms and the ${4x}$ terms cancel:

$\begin{align*}&f(x+h)-f(x)\\&=(x+h)^2-4(x+h)-(x^2-4x)\\&=x^2+2xh+h^2-4x-4h-x^2+4x\\&=2xh+h^2-4h\end{align*}$

So

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\dfrac{2xh+h^2-4h}{h}\end{align*}$

Every term in the numerator has a factor ${h}$. Cancel it, and let ${h}$ approach ${0}$:

$\begin{align*}&=\lim_{h\to 0}(2x+h-4)\\&=2x-4\end{align*}$

The tangent line is horizontal where its slope is ${0}$:

$\begin{align*}2x-4&=0\\x&=2\end{align*}$

So the tangent line is horizontal at ${x=2}$.

Use the definition of the derivative:

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\dfrac{\dfrac{1}{x+h+2}-\dfrac{1}{x+2}}{h}\end{align*}$

Combine the two fractions in the numerator, using the common denominator ${(x+h+2)(x+2)}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{\dfrac{(x+2)-(x+h+2)}{(x+h+2)(x+2)}}{h}\end{align*}$

Simplify the numerator. Dividing by ${h}$ puts ${h}$ in the denominator:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-h}{h(x+h+2)(x+2)}\end{align*}$

Cancel the common factor ${h}$, and let ${h}$ approach ${0}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-1}{(x+h+2)(x+2)}\\&=-\dfrac{1}{(x+2)^2}\end{align*}$

This is defined for every ${x\ne -2}$.

Use the definition of the derivative:

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\dfrac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\end{align*}$

Direct substitution gives ${0/0}$. So multiply the numerator and the denominator by the conjugate, ${\sqrt{x+h+1}+\sqrt{x+1}}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{(x+h+1)-(x+1)}{h(\sqrt{x+h+1}+\sqrt{x+1})}\\&=\lim_{h\to 0}\dfrac{h}{h(\sqrt{x+h+1}+\sqrt{x+1})}\end{align*}$

Cancel the common factor ${h}$, and let ${h}$ approach ${0}$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{1}{\sqrt{x+h+1}+\sqrt{x+1}}\\&=\dfrac{1}{2\sqrt{x+1}}\end{align*}$

The function ${f}$ is defined for ${x\ge -1}$. But ${f'(x)}$ is defined only for ${x>-1}$, because the denominator ${2\sqrt{x+1}}$ is ${0}$ when ${x=-1}$. So the domain of ${f'}$ is ${(-1,\infty)}$.

Here ${f(3)=0}$ and ${f(3+h)=|h|}$. So the difference quotient is

${\dfrac{f(3+h)-f(3)}{h}=\dfrac{|h|}{h}}$

For ${h>0}$, ${|h|=h}$, so the quotient is ${1}$. For ${h<0}$, ${|h|=-h}$, so the quotient is ${-1}$:

$\displaystyle\lim_{h\to 0^+}\dfrac{|h|}{h}=1$   and   $\displaystyle\lim_{h\to 0^-}\dfrac{|h|}{h}=-1$

The one-sided limits are different, so the limit does not exist. So ${f'(3)}$ does not exist. On the graph, ${y=|x-3|}$ has a sharp corner at ${(3,0)}$.

False. The function in part (a), ${f(x)=|x-3|}$, is continuous at ${3}$. But it is not differentiable at ${3}$, because the graph has a corner there.

True. This is the theorem "differentiable functions are continuous." If ${f'(a)}$ exists, then ${f(a+h)}$ approaches ${f(a)}$ as ${h\to 0}$. So ${f}$ is continuous at ${a}$.