to get the slope of the tangents at those points: − and −2, respectively. The slopes of the
normal lines are the opposite reciprocals of those slopes, so and , respectively. The
equations for the normal lines are then found by using the point-slope formula. The equations
are y + 2 = (x + 1) and y − 1 = (x + 1).
- Let f be the function given by f(x) = 3x^3 − 6x^2 + 4x.
(a) Find an equation for the normal line at x = 2.
First, determine what f(2) equals: f(2) = 8. Next, take the first derivative of f(x): f′(x) = 9x^2 −
12 x + 4. At x = 2, f′(x) = 16. The slope of the normal line would then be − . The equation of
the normal line is y − 8 = − (x − 2).
(b) Where are the relative maxima and minima of the curve, if any exist? Verify your answer.
Relative maxima and minima exist where the first derivative is zero or does not exist. Set the
first derivative equal to zero and solve or x, f′(x) = 0 at x = . To determine whether this point
is a maximum or minimum, check the second derivative. If the second derivative is negative,
the point is a maximum. If the second derivative is positive, the point is a minimum. However,
if the second derivative is zero, the point is a point of inflection. At x = , f′ (x) = 0, so the
point is a point of inflection. This curve has no maxima or minima.
(c) Where are the points of inflection? Verify your answer. If there are none, explain why.
It was discovered in part (b) that the point of inflection is at x = . Plug that into f(x) to find
the corresponding y-value, which is . Thus, the point of inflection is located at .
Justification is the same as the justification in part (b).
