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Find the tangent line to the graph of a cubic at a particular point

by RoRi

Consider the cubic curve

    \[ f(x) = x - x^3. \]

Find values for m and b such that the line y = mx + b is tangent to the graph of f at the point (-1,0).

There is a second line passing through the point (-1,0) which is tangent to the graph of f at the point (a,c). Find the values of a and c.

First, we sketch the graph of f on the interval [-2,2]:

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We compute the derivative of f,

    \[ f'(x) = 1 - 3x^2. \]

So, the slope of the tangent line at the point (-1,0) is

    \[ f'(-1) = 1 - 3(-1)^2 = -2 \quad \implies \quad m = -2. \]

Then, since y = mx + b is the tangent line to the curve at (-1,0) it must be on the curve at this point, so

    \[ y(-1) = 0 \quad \implies \quad (-2)(-1) + b = 0 \quad \implies \quad b = -2. \]

Therefore, the line is y = -2x -2, i.e., m = b = -2.

Next, if there is another line, say y_1 = m_1 x + b_1, tangent to f at a point (a,c) we know it has slope given by

    \[ f'(a) = 1 - 3a^2 = m_1. \]

Since it must pass through the point (-1,0) (by hypothesis) we must have

    \[ y_1(-1) = 0 \quad \implies \quad (1-3a^2)(-1) + b_1 = 0 \quad \implies \quad b_1 = 1 -3a^2. \]

So the line y_1 is of the form,

    \[ y_1 = (1 - 3a^2)x + (1-3a^2). \]

Finally, since the point (a,c) is on both this line y_1 and the curve f we have,

    \begin{align*} f(a) &= c & \text{and} && y_1(a) &= c \\ a-a^3 &=c & \text{and} && (1-3a^2)(a) + (1-3a^2) &=c. \end{align*}

Since these are both equal to c, they must be equal to each other,

    \begin{align*} &&a - a^3 &= (1-3a^2)(a) + (1-3a^2) \\ \implies && 2a^3 + 3a^2 - 1 &= 0 \\ \implies && a &= \frac{1}{2}. \end{align*}

Since a - a^3 = c we then have c = \frac{3}{8}.

Thus, a = \frac{1}{2}, and c = \frac{3}{8} are the requested values.

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