2.2 The Tangent Line Problem

 

2.2.1 Finding the Slope of a Tangent Line

 

Goals

 

In this section, we solve the tangent line problem for polynomial functions. As a byproduct, we also solve the tangent line problem for rational functions, and for root functions. We prove the following basic rules of differentiation. Let \(n, m \in \mathbb{N}\), then

(1) if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\);

(2) if \(f(x) = x^{-n}\), \(x\neq 0\), then \(f'(x) = -nx^{-n-1}\);

(3) if \(\displaystyle f(x) =\sqrt[n]{x^m} = x^{m/n}\), then \(\displaystyle f'(x) = -\frac{m}{n}x^{\frac{m}{n} – 1}\), provided \(f\) and \(f’\) are defined at \(x\).

 

Tangent Line Problem and Approximation Method

 

Given a function \(f\) and a point \(a\in D_f\), find an equation for the line \(\mathcal{L}_T\) tangent to the graph at \((a,f(a))\). To solve this problem, we need to the slope \(m_T\) of the tangent line at the point.

The method to find the slope of the tangent line—if such a tangent line exists—consists on approximating the slope of the tangent line at \((a,f(a))\) by means of slopes of secant lines \(m_S\) to the graph of a function \(f\), passing through \((a,f(a))\):
\[m_S= \frac{f(x)-f(a)}{x-a}, \;\;\;\; x\neq a.\]

The sequence of pictures in Figure~\ref{ch02-02-f01}, shows that as \(x\) approaches \(a\) the secant line approaches the tangent line. Or, the slopes of the secant lines approach the slope of the tangent line.

The notation to represent the approximation process to the slope of the tangent line is as follows:
\[m_T=\lim_{x\to a} m_S,\]
and, using the difference quotient of \(f\),
\[
m_{T}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\;\;\;\; \text{or} \;\;\;\; m_{T}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.
\]
This notation emphasizes the fact that we are looking at the values of the slopes of the secant lines as \(x\) approaches \(a\), \(x\neq a\), with the aim to see if they approach a value that yields the slope of the tangent line. It is read as follows, respectively,

“the limit as \(\;x\;\) approaches \(\;a\;\) of \(\displaystyle \;\frac{f(x)-f(a)}{x-a}\;\) is \(\;m_T\),”

“the limit as \(\;h\;\) approaches \(\;0\;\) of \(\displaystyle\; \frac{f(a+h)-f(a)}{h}\;\) is \(\;m_T\).”

These limits, when they exist, yield the slope of the tangent line to the graph of \(f\) at \((a,f(a))\). The limits (if they exist) are the derivative of \(\pmb{f}\) at \(\pmb{a}\), and is denoted by \(f'(a)\). Thus the following definitions. (We will come back to this in the next section.)

 

Definition of Derivative and Direct Substitution Property

 

Definition 1. Let \(f\) be a function defined on an open interval containing \(A\). The slope of the tangent line \(m_T\) to the graph of \(f\) at \((a,f(a))\) is
\[m_{T}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}
\;\;\;\; \text{or} \;\;\;\;
m_{T} =\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},\]
provided these limits exist.

Definition 2. Let \(f\) be a function defined on an open interval containing \(a\). The derivative of \(f\) at \(a\) is
\[f'(a) =\lim_{x\to a}\frac{f(x)-f(a)}{x-a},
\;\;\; \text{or} \;\;\;
f'(a) =\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},\]
provided these limits exist.

To compute these limits, we use the following

Direct Substitution Property (DSP). If \(f\) is a polynomial, rational, or root function, then
\[\lim_{x\to a}f(x)=f(a), \;\;\; \text{for all } \; a \in D_f.\]
Namely, for polynomial or rational functions, the limit as \(x\) approaches \(a\) is the value of the function at \(a\), \(f(a)\), for \(a\in D_f\).

 

We are now ready to find tangent lines to graphs of polynomial and rational functions.

 

Example 1. Given the function \(f(x)=x^3-4x+5\), find the slope-intercept equation of the tangent line to the graph of \(f\) at the point \((2,f(2))\).

 

Solution. First, we compute the difference quotient. Let \( x \in \mathbb{R}\) with \(x \neq 2\). Then,
\[ \frac{f(x)-f(2)}{x-2} = \frac{x^3-4x+5-\left(2^3-4\cdot 2+5\right)}{x-2} = \frac{x^3 – 2^3 }{x-2}- \frac{ -4(x-2)}{x-2} \underset{?}{=}x^2+2x.\]
We now apply the DSP:
\[m = \lim_{x\to 2} \frac{f(x)-f(2)}{x-2}= \lim_{x\to 2}( x^2+2x) \underset{\text{(DSP)}}{=}2^2+2(2)= 8.\]

Since we are looking for the tangent line at the point \((2,f(2))= (2, 5)\), we can use the point-slope equation of the line:
\[y-5=8(x-2)\;\; \Longleftrightarrow \;\; y = 8x-11.\]

Example 2. Given the function \(\displaystyle f(x)=\sqrt[5]{x^3}\), find the slope-intercept equation of the tangent line to the graph of \(f\) at the point with \(x\)-coordinate \(x=-1\).

Solution.
Since in Example 3 in 2.1.3 it was proven that
\[\frac{\sqrt[5]{x^3} – \sqrt[5]{a^3}}{x-a} =\frac{\sqrt[5]{x^2} +\sqrt[5]{xa} + \sqrt[5]{a^2}}{\sqrt[5]{x^4}+\sqrt[5]{x^3a} +\sqrt[5]{x^2 a^2}+\sqrt[5]{x a^3}+\sqrt[5]{a^4}}, \;\; x\neq a,\]
we can use this result, and substitute \(x=-1\):
\begin{align*}
\frac{\sqrt[5]{x^3} – \sqrt[5]{(-1)^3}}{x-(-1)} &=\frac{\sqrt[5]{x^2} +\sqrt[5]{x(-1)} + \sqrt[5]{(-1)^2}}{\sqrt[5]{x^4}+\sqrt[5]{x^3 (-1)}+\sqrt[5]{x^2 (-1)^2}+\sqrt[5]{x (-1)^3}+\sqrt[5]{(-1)^4}} =\\
&=\frac{\sqrt[5]{x^2} -\sqrt[5]{x} + 1}{\sqrt[5]{x^4}-\sqrt[5]{x^3}+\sqrt[5]{x^2} – \sqrt[5]{x}+1}, \;\; x\neq -1,
\end{align*}
We can now apply the DSP:
\begin{align*}
m_T = \lim_{x\to -1} \frac{\sqrt[5]{x^3} +1}{x+1}&\underset{\hphantom{\text{(DSP)}}}{=} \lim_{x\to -1} \frac{\sqrt[5]{x^2} -\sqrt[5]{x} + 1}{\sqrt[5]{x^4}-\sqrt[5]{x^3}+\sqrt[5]{x^2} – \sqrt[5]{x}+1} =\\
&\underset{\text{(DSP)}}{=} \frac{\sqrt[5]{(-1)^2} -\sqrt[5]{(-1)} + 1}{\sqrt[5]{(-1)^4}-\sqrt[5]{(-1)^3}+\sqrt[5]{(-1)^2} – \sqrt[5]{(-1)}+1} = \frac{3}{5}.
\end{align*}
Since we are looking for the tangent line at the point \((-1,f(-1))= (-1, -1)\),
\[y+1=\frac{3}{5}(x+1) \Longleftrightarrow y =\frac{3}{5} x -\frac{2}{5}.\]

Example 3. If \( g(14) = 12\) and \(\displaystyle g'(14) = \frac{4}{3}\), find \(x_2\) such that the point \( B\left(x_2, -12\right)\) is on the line tangent to the graph of \(y = g(x)\) at the point \( A\left(14, g(14)\right)\).

Solution.

(1) Since both \(A\) and \(B\) lie on the tangent line, its slope is given by \(\displaystyle m = \frac{-12-12}{x_2-14}\).

(2) The slope of the tangent line is also given by the derivative, that is, \(\displaystyle m=g'(14)=\frac{4}{3}\).

(3) Hence \(\displaystyle \frac{4}{3} = \frac{-12-12}{x_2-14}\). Solving for \(x_2\), we have \(x_2=-4\).

 

Basic Differentiation Formulas

 

Proposition 1. Let \( f(x) = x^n\), \(n\in \mathbb{N}\), then the slope of the line tangent to its graph at a point with \(x\)-coordinate \(a \in \mathbb{R}\) is given by \(f'(a)=na^{n-1}\).

Proof.

\begin{align*}
f'(a)&\underset{\hphantom{(\text{DSP)}}}{=}\lim_{x \to a} \frac{ x^n – a^n}{x-a}
\underset{(\ref{ch02-eq04})}{=}\lim_{x \to a}( x^{n-1} + x^{n-2}a + x^{n-3}a^2 +\cdots x^2a^{n-3} +x a^{n-2}+ a^{n-1}) =\\
&\underset{(\text{DSP})}{=} a^{n-1} + a^{n-2}a + a^{n-3}a^2 +\cdots a^2a^{n-3} +a a^{n-2}+ a^{n-1}
= n a^{n-1}.
\end{align*}
Note that there are exactly \(n\) terms in the sum.

 
Corollary 1. Let \( f(x) = x^{-n}\), \(n\in \mathbb{N}\), then the slope of the line tangent to its graph of \(f\) at a point with \(x\)-coordinate \(a \in \mathbb{R}\), \(a\neq 0\), is given by \(f'(a)=-na^{-n-1}\).

The proof is left as an exercise (see #16(a)).
 

Combining Proposition 1 and Corollary 1, we have the following general formula:
\begin{equation}\label{ch02-eq10}
f(x)=x^n \;\; \Longrightarrow \;\; f'(a)= na^{n-1}, \;\; n\in \mathbb{Z}, \; a\in D_f .\tag{2.10}
\end{equation}

 

Proposition 2. Let \( f(x) = \sqrt[n]{x^{m}}\), \(n, m\in \mathbb{N}\), \(n \geq 2\). Then
\begin{equation}\label{ch02-eq11}
f(x)=x^{\frac{m}{n}} \;\; \Longrightarrow \;\; f'(a)= \frac{m}{n} a^{\frac{m}{n}-1}, \;\; \; a\in D_f\cap D_f’ .\tag{2.11}
\end{equation}

The proof is left as an exercise (see #16(b)).

Example 4. Given \( f(x) = 3x^5-4x^3+4\), find the slope-intercept equation of the line tangent to the graph of \(f\) at the point with \(x=-1\).

Solution. To find the slope of the tangent line at \(x=-1\), we need to compute \(f'(-1\). Since
\[f'(a) = 15a^4-12a^3 \;\; \Longrightarrow \;\; f'(-1) = 15(-1)^4-12(-1)^2=3.\]
Hence, we have

• The point on the graph has coordinates \((-1,f(-1))=(-1,5)\).
• The slope of the tangent line at \((-1,5)\) is given by \(f'(-1)=3\).
• Point-slope equation of the tangent line: \(y-5=3(x-(-1))\).
• Slope-intercept equation of the tangent line: \(y= 3 x + 8\).

 

2.2.3 Problem Solving: Tangent Lines

The following problem was solved using algebraic procedures in Example~2 in 1.2.4.

Example 1. Given the parabola \(\mathcal{P}\) with equation \(y=2 x^2-4 x+3\), and the secant line \(\mathcal{L}_S\) passing through the points \(A\) and \(B\) with \(x\)-coordinates, respectively, \(x=0\) and \(x=4\), find the point on \(\mathcal{P}\) whose tangent line is parallel to \(\mathcal{L}_S\), and find the slope-intercept equation of such tangent line.

Solution. \(\mathcal{L}_S\) passes through the points with coordinates

\(A(0, f(0))=(0,3)\) and \(B(4, f(4))= (4, 19)\) (verify this).
Thus, \(\mathcal{L}_S\) has slope
\[m_S=\frac{19-3}{4-0}=4,\]
which is also the slope of \(\mathcal{L}_T\) (since \(\mathcal{L}_S \, \parallel\, \mathcal{L}_T\)).

 

Let \(P(x,y)\) be the point on \(\mathcal{P}\) where the tangent line \(\mathcal{L}_T\) is parallel to \(\mathcal{L}_S\).

 

(1) The slope of \(\mathcal{L}_T\) is given by the derivative, that is \(y’=4x-4\) (verify this).

 

(2) Since \(\mathcal{L}_T\) must be parallel to \(\mathcal{L}_S\), both lines must have equal slopes. Namely,
\[y’= m_S \Longleftrightarrow 4x-4 = 4 \Longleftrightarrow x= 2 \Longrightarrow y \underset{?}{=} 8-8+3= 3 \Longrightarrow P(2,3). \]
Thus, the point-slope equation of \(\mathcal{L}_T\), and the slope-intercept equation are given by, respectively,
\[y-3=4(x-2) \Longleftrightarrow y = 4x-5.\]

Example 2. Find the points on the graph of \(f(x)= x^4-x^3\) whose tangent line is horizontal.

Solution. The slope of the tangent line at a point with \(x\)-coordinate \(x=a\) is given by \(f'(a)=4a^3-3a^2\). Further, the tangent line is horizontal if its slope is zero, that is,
\[f'(a)=0\Longleftrightarrow 4a^3-3a^2=0 \Longleftrightarrow a^2(4a-3)=0 \Longleftrightarrow a=0 \;\; \text{or} \;\; a=\frac{3}{4}.\]
It follows that the graph has two points whose tangent line is horizontal: \[(0, f(0))=(0,0) \text{ and } \left (\frac{3}{4}, f\left (\frac{3}{4}\right )\right )=\left (\frac{3}{4}, -\frac{27}{256}\right ).\]

The following example is similar to Example 1. However, in this case it is not immediate that the problem can be solved algebraically. Thus, we use the derivative to solve it.

Example 3. Given \(g(x)=x^3+3 x^2+x-1\), and the secant line \(\mathcal{L}_S\) passing through the points \(A\) and \(B\) with \(x\)-coordinates, respectively, \(x=0\) and \(x=-3\), find the point on the graph of \(g\) whose tangent lines are parallel to \(\mathcal{L}_S\), and find the slope-intercept equation of such tangent lines.

Solution. \(\mathcal{L}_S\) passes through the points with coordinates

\(A(0, g(0))=(0,-1)\) and \(B(-3, g(-3))= (-3, -4)\) (verify this).
Thus, \(\mathcal{L}_S\) has slope \(m_S=1\) (verify this) which is also the slope of \(\mathcal{L}_T\) (since \(\mathcal{L}_S \, \parallel\, \mathcal{L}_T\)).

 

Let \(P(x,y)\) be a point on the graph of \(g\) where the tangent line \(\mathcal{L}_T\) is parallel to \(\mathcal{L}_S\).

 

(1) The slope of \(\mathcal{L}_T\) is given by the derivative, that is \(g'(x)=3x^2+6x+1\) (verify this).

 

(2) Since \(\mathcal{L}_T\) must be parallel to \(\mathcal{L}_S\), both lines must have equal slopes. Namely,
\[g'(x)= m_S \Longleftrightarrow 3x^2+6x+1 = 1. \]
It follows that \(P(x,y)\) has \(x\)-coordinate \(x=-2\) or \(x=0\) (verify this), and hence \(y =1\) or \(y=0\), respectively (verify). That is, \(P_1(-2,1)\) and \(P_2(0,0)\) are the points whose tangent line is parallel to \(\mathcal{L}_S\). Thus, the point-slope equation of the corresponding tangent lines are given, respectively by (verify)
\[y = x+3 \;\; \text{and}\;\; y =x-1.\]

Example 4. The graph of the quadratic function \( f(x)= x^2+b x +c\) passes through the point \( P_0(-1,4)\), and its tangent line at \(P_0\) is parallel to \( 3x + y = 4\).

(a) find the coefficients \(b\) and \(c\).

(b) find the \(x\)-coordinate of the point on the graph of \(f\) whose tangent line is perpendicular to \(-2x+y=2\).

Solution. (a) We need two equations.

• The graph of \( f(x)= x^2+b x+c\) passes through the point \( P_0(-1,4)\):
  \[f(-1)=4\Longleftrightarrow 1-b+c=4 \Longleftrightarrow b-c=-3\,.\hspace{1in}(*)\]
• The tangent line at \(P_0\) is parallel to \( 3x + y = 4\): \(f'(-1)=-3 \Longleftrightarrow -2+b=-3\Longleftrightarrow b=-1\).
• Then, substituting \(b=-1\) in \((*)\): \(1-(-1)+c=4\Longleftrightarrow c=2\).
• Thus, \(f(x)=x^2-x+2\).

(b) To find the \(x\)-coordinate of the point on the graph of \(f\) whose tangent line is perpendicular to \( -2x+y=2\), we need to solve
\[f'(x)\overset{?}{=}-\frac{1}{2} \Longleftrightarrow 2x-1= -\frac{1}{2} \Longleftrightarrow x=\frac{1}{4}\,.\]

Example 5. Given, \(\displaystyle f(x)=\frac{1}{2}x(3-x)\), find the area of the triangle formed by the coordinate axes and the tangent line to the graph of \(f\) at the point with \(x\)-coordinate \(x=2\) (see Figure 2.3(a)).

Solution. (1) Equation of the line tangent to the graph at the indicated point.

• Point of tangency: \(P(2, f(2))=(2, 1)\)
• Slope of the tangent line at \(P\): \(m=f'(2)\). Since \(\displaystyle f'(x)= \frac{3}{2}-x\), \(m=-\frac{1}{2}\).
• Point-slope equation for the tangent line: \(\displaystyle y-1=-\frac{1}{2}(x-2)\).
• Slope-intercept equation of the tangent line: \(\displaystyle y=-\frac{1}{2}x+2\).

(2) \(x\)- and \(y\)-intercepts of the tangent line: \(x\)-intercept \((4,0)\), \(y\)-intercept \((0,2)\).

 

(3) Area of the triangle: \(\displaystyle A=\frac{1}{2}(4)(2)=4\).

Example 6. Find the points on the parabola \( \displaystyle y =x^2-4 x+6\) whose tangent lines pass through the point \( \displaystyle A\left(3,-1\right)\) (see Figure 2.3(b)).

Solution. Let \(P(x,y)\) be such a point. To determine the coordinates of \(P\), we need two equations. One equation is readily available: the coordinates of \(P\) must satisfy the equation of the parabola. Thus, \( y =x^2-4 x+6\).

To find a second equation, we compute the slope, \(m\), of the line tangent to the parabola at \(P\) in two different ways:

• On one hand, \(\displaystyle m = \frac{y-(-1)}{x-3}\) (explain why).
• On the other hand, \(m =f'(x)=2x-4\) (explain why).

Since these are two different expressions of the same slope, they must be equal. Thus, a second equation is given by
\[\frac{y-(-1)}{x-3}= 2x-4 \;\; \Longleftrightarrow \;\;
\left \{\renewcommand{\arraystretch}{1.2}\begin{array}{l}
y+1 = (x-3)(2x-4)\\ x\neq 3
\end{array} \right .
\;\; \Longleftrightarrow \;\;
\left \{\renewcommand{\arraystretch}{1.2}\begin{array}{l}
y = 2 x^2-10 x+11\\x\neq 3
\end{array} \right .
.\]
We now solve simultaneously the two equations
\[y =x^2-4 x+6 \;\;\; \text{ and } \;\;\;y = 2 x^2-10 x+11.\]
\begin{align*}
\left \{\renewcommand{\arraystretch}{1.2}\begin{array}{l}
y =x^2-4 x+6 \\y = 2 x^2-10 x+11
\end{array} \right.&\;\;\; \Longrightarrow \;\;
x^2-4 x+6 = 2 x^2-10 x+11 \;\; \Longleftrightarrow \;\; x^2-6 x+5 =0 \;\; \Longleftrightarrow\\
&\;\; \Longleftrightarrow\;\; (x-5) (x-1)=0 \;\; \Longleftrightarrow x\in \left \{1,5\right \}.
\end{align*}
We leave to the reader to verify that the points on the parabola are \(\left \{(1,3), (5,11)\right \}\).