Chapter 2.6, Problem 3E

(a)

To determine

To find: The slope of the tangent line to the parabola $y=4x-x^2$ at the point.

Answer to Problem 3E

The slope of the tangent line to the parabola $y=4x-x^2$ at the point $\left(1,3\right)$,

(i) Using Definition 1 is $m=2$.

(ii) Using Equation 2 is $m=2$.

Explanation of Solution

Given:

The slope of the tangent line to the parabola $y=4x-x^2$ at the point $\left(1,3\right)$.

Formula used:

Definition 1: The slope of the tangent curve $y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$, $m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

Equation 2: The slope of the tangent line in definition 1 becomes, $m=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

Section (i)

Obtain the slope of the tangent line to the parabola at the point (1, 3) by using Definition 1.

Substitute $a=1$ and $f\left(a\right)=3$ in Definition 1,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{\left(4x-x^2\right)-3}{x-1}\\ =\underset{x\to 1}{\lim }\frac{4x-x^2-3}{x-1}\\ =\underset{x\to 1}{\lim }\frac{-\left(x^2-4x+3\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{-\left(x^2-3x-x+3\right)}{x-1}\end{array}$

The factors of $\left(x^2-4x+3\right)$ are $\left(x-3\right)$ and $\left(x-1\right)$.

Therefore, the slope of the tangent line to the parabola becomes,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{-\left(x\left(x-3\right)-\left(x-3\right)\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{-\left(x-1\right)\left(x-3\right)}{\left(x-1\right)}\end{array}$

Since the limit x approaches 1 but not equal to 1, cancel the common term $\left(x-1\right)\left(\ne 0\right)$ from both numerator and the denominator,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\left(-\left(x-3\right)\right)\\ =\underset{x\to 1}{\lim }\left(3-x\right)\\ =3-1\\ =2\end{array}$

Thus, the slope of the tangent line to the parabola at the point (1, 3) is $\underset{\_}{m=2}$.

Section (ii)

Obtain the slope of the tangent line to the parabola at the point (1, 3) by using Equation 2.

Substitute $a=1$ and $f\left(a\right)=3$ in Equation 2,

$\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(4\left(1+h\right)-{\left(1+h\right)}^2\right)-3}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(4+4h-\left(1+h^2+2h\right)\right)-3}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(4+4h-1-h^2-2h\right)-3}{h}\end{array}$

Simplify further and obtain the value of m.

$\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{\left(3+2h-h^2\right)-3}{h}\\ =\underset{h\to 0}{\lim }\frac{3+2h-h^2-3}{h}\\ =\underset{h\to 0}{\lim }\frac{2h-h^2}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(2-h\right)}{h}\end{array}$

Since the limit h tends to 0 but not equal to 0, cancel the common term $h\left(\ne 0\right)$ from both the numerator and the denominator,

Thus, the slope of the tangent line becomes,

$\begin{array}{c}m=\underset{h\to 0}{\lim }\left(2-h\right)\\ =2-0\\ =2\end{array}$

Thus, the slope of the tangent line to the parabola at the point (1, 3) by using equation 2 is $\underset{\_}{m=2}$.

(b)

To determine

To find: The equation of the tangent line in part(a).

Answer to Problem 3E

The equation of the tangent line in part (a) is $\underset{\_}{y=2x+1}$.

Explanation of Solution

Equation of the tangent line:

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is,

$y-f\left(a\right)=f^{\prime }\left(a\right)\left(x-a\right)$

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of a, the value of $f^{\prime }\left(a\right)=2$.

Substitute $a=1$, $f\left(a\right)=3$ and $f^{\prime }\left(a\right)=2$ in the equation of tangent line,

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-3=2\left(x-1\right)\\ =2x-2\\ \end{array}$

Isolate y as shown below.

$\begin{array}{c}y=2x-2+3\\ =2x+1\end{array}$

Thus, the equation of the tangent line is $\underset{\_}{y=2x+1}$.

(c)

To determine

To sketch: The graph of the function is the tangent line, $y=2x+1$ and the parabola, $y=4x-x^2$ and check zoom in the graph toward the point (1, 3) until the parabola and the tangent line is indistinguishable.

Explanation of Solution

Calculation:

The equation of the tangent line is $y=2x+1$.

The equation of the given curve is $y=4x-x^2$.

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

Graph:

Use the online graphing calculator to zoom toward the point (1,3) as shown below in Figure 2.

From Figure 2, it is observed that when zoom in the graph toward the point (1, 3), the graph of the tangent line and the curve looks likes almost identical.

Hence, it is verified that the graph of the tangent line and the parabola zoom in toward the point (1, 3) until the tangent line are indistinguishable.