Equation of Tangent Line

Question

Use the limit derivative to compute the derivative of the function f(t) = 5/4-t at t = -3

f’(-3) = 5/49

Find an equation of the tangent line to the graph of f at t = -3.

What is the equation?

Answer 1

Assuming f(t) is of the form f(t) = 5/(4 - t), and using the limit definition of the derivative which states:

f'(t) = lim_h->0 (f(t + h) - f(t))/h

We can do a few computations followed by a substitution. First,

f(t + h) = 5/(4 - (t + h)) = 5/(4 - t - h)

Next,

f(t + h) - f(t) = 5/(4 - t - h) - 5/(4 - t)

f(t + h) - f(t) = (5(4 - t) - 5(4 - t - h))/((4 - t)(4 - t - h))

f(t + h) - f(t) = (20 - 5t - 20 + 5t + 5h)/(16 - 4t - 4h - 4t + t² + th)

f(t + h) - f(t) = 5h/(16 - 8t - 4h + t² + th)

Now, dividing by h yields

(f(t + h) - f(t))/h = 5/(16 - 8t - 4h + t² + th)

Finally, we take the limit as h -> 0 such that:

f'(t) = lim_h->0 (f(t + h) - f(t))/h

f'(t) = lim_h->0 5/(16 - 8t - 4h + t² + th)

f'(t) = 5/(16 - 8t + t²)

f'(t) = 5/(4 - t)²

As you have stated, f'(-3) = 5/(4 - (-3))² = 5/49. This tells us that the slope of the tangent line is equal to 5/49 so we can write the tangent line as y = 5x/49 + b where b is the yet to be determined y-intercept.

Since we are looking at the function at x = -3, we can compute f(-3) = 5/(4 - (-3)) = 5/7. This tells us that (-3, 5/7) is a point on the graph of the function and is thus a point on the tangent line (the only point that they have in common, actually). We can return to our tangent line, knowing now that the point (-3, 5/7) is a solution to the equation of the line. Substitution yields:

5/7 = 5(-3)/49 + b

35/49 = -15/49 + b

50/49 = b

Thus, the equation of the tangent line is y = 5x/49 + 50/49.