Worked examples · Sheet 3

Tangent lines

Theory used: Tangent line to the graph Derivative of sum, product and quotient Derivatives of the elementary functions

Exercise 1

Write the equation of the tangent line to the graph of $f (x) = x^{3} - 2 x$ at the point with abscissa $x_{0} = 1$ .

f (1) = 1 - 2 = - 1

f^{'} (x) = 3 x^{2} - 2, f^{'} (1) = 3 - 2 = 1

y = f (1) + f^{'} (1) (x - 1) = - 1 + (x - 1) = x - 2

The tangent line is $y = x - 2$ .

The tangent line passes through the point $(x_{0}, f (x_{0}))$ and has slope $f^{'} (x_{0})$ , therefore the two values $f (1)$ and $f^{'} (1)$ are needed. The value of the function is $f (1) = - 1$ . The derivative is computed with the sum and power rules and equals $3 x^{2} - 2$ , hence the slope is $f^{'} (1) = 1$ . Substituting into the tangent line formula one obtains $y = - 1 + 1 \cdot (x - 1)$ , that is, the line $y = x - 2$ .

Exercise 2

Find the points of the graph of $f (x) = x^{3} - 3 x$ at which the tangent line is horizontal and write the corresponding tangents.

f^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1) = 0 ⟺ x = - 1 or x = 1

f (- 1) = - 1 + 3 = 2, f (1) = 1 - 3 = - 2

The points are $(- 1, 2)$ and $(1, - 2)$ ; the tangents are the lines $y = 2$ and $y = - 2$ .

y = x^{3} - 3 x

y = x^{3} - 3 x

: at the points

(- 1, 2)

and

(1, - 2)

the tangents

y = 2

and

y = - 2

are horizontal

A line is horizontal when its slope is zero, therefore one looks for the points where $f^{'} (x) = 0$ . The derivative equals $3 x^{2} - 3$ and vanishes for $x^{2} = 1$ , that is, at the two points $x = - 1$ and $x = 1$ . The corresponding ordinates are $f (- 1) = 2$ and $f (1) = - 2$ . At a point with a horizontal tangent the tangent is the constant line through the point, hence the two tangents are $y = 2$ and $y = - 2$ .

Exercise 3

Find the point of the graph of $f (x) = ln x$ at which the tangent line is parallel to the line $y = \frac{x}{2}$ and write the tangent.

f^{'} (x) = \frac{1}{x} = \frac{1}{2} ⟺ x = 2

f (2) = ln 2

y = ln 2 + \frac{1}{2} (x - 2) = \frac{x}{2} + ln 2 - 1

The point is $(2, ln 2)$ ; the tangent is the line $y = \frac{x}{2} + ln 2 - 1$ .

y = \ln x

y = \dfrac{x}{2}

y = \dfrac{x}{2} + \ln 2 - 1

The tangent to

y = ln x

at the point

(2, ln 2)

(red) is parallel to the line

y = \frac{x}{2}

(dashed)

Two lines are parallel when they have the same slope, therefore one looks for the point where the derivative equals $\frac{1}{2}$ , the slope of the given line. The derivative of the logarithm is $\frac{1}{x}$ and the equation $\frac{1}{x} = \frac{1}{2}$ has the unique solution $x = 2$ , which belongs to the domain $(0, + \infty)$ . The ordinate of the point of tangency is $f (2) = ln 2$ . The tangent passes through $(2, ln 2)$ with slope $\frac{1}{2}$ , hence its equation is $y = \frac{x}{2} + ln 2 - 1$ , parallel to the given line and distinct from it.