Question

Find an equation of a line that is tangent to the graph of f and parallel to the given line.Function: f(x) = x^3 Line: 3x − y + 4 = 0

Answer 1

SOLUTION

From the line

$\begin{gathered} 3x-y+4=0 \\ -y=-3x-4 \\ \text{dividing by -1} \\ \frac{-y}{-1}=\frac{-3x}{-1}+\frac{-4}{-1} \\ y=3x+4 \end{gathered}$

Equation of the line parallel to this line must have the same slope of the line.

From equation of a line in slope-intercept form,

$\begin{gathered} y=mx+c \\ \text{where m is the slope and c the intercept on the y-axis, then } \\ y=3x+4\text{ has a slope of 3,} \\ \text{that is m = 3} \end{gathered}$

Now, let the new line have the equation

$\begin{gathered} y=3x+c,\text{ since it has the same slope (3) of the other line } \\ So,\text{ now, we have to find c} \end{gathered}$

From the function

$\begin{gathered} f(x)=x^3 \\ f^{\prime}(x)=3x^2 \\ To\text{ be tangent to the line }y=3x+c,\text{ the curve should have same } \\ \text{slope and be equal to the line, hence } \\ 3x^2=3 \\ x^2=1 \\ x=\sqrt[]{1} \\ x=1 \end{gathered}$

Substituting the value of x into f(x), we have

$\begin{gathered} f(x)=x^3 \\ y=1^3 \\ =1 \end{gathered}$

So, they are both tangent at (1, 1).

From the equation above we have

$y=3x+c$

Substituting the values of x = 1 and y = 1 in the above equation, we have

$\begin{gathered} y=3x+c \\ 1=3(1)+c \\ 1=3+c \\ 1-3=c \\ -2=c \\ c=-2 \end{gathered}$

Substituting the c for -2 back into the original equation we have

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

Hence the answer is

$y=3x-2$