Question

Consider the equation below. f(x) = 2x^3 + 3x^2 − 12x (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) Find the interval on which f is decreasing. (Enter your answer in interval notation.) (b) Find the local minimum and maximum values of f. local minimum local maximum (c) Find the inflection point. (x, y) = Find the interval on which f is concave up. (Enter your answer in interval notation.) Find the interval on which f is concave down. (Enter your answer in interval notation.)

Answer 1

Answer:

a) increasing: (-∞, -2)∪(1, ∞); decreasing: (-2, 1)

b) local maximum: (-2, 20); local minimum: (1, -7)

c) inflection point: (-0.5, 6.5); concave up: (-0.5, ∞); concave down: (-∞, 0.5)

Step-by-step explanation:

A graphing calculator can show you the local extremes. Everything else falls out from those.

a) The leading coefficient is positive, so the general shape of the graph is from lower left to upper right. The function will be increasing from -infinity to the local maximum (x=-2), decreasing from there to the local minimum (x=1), then increasing again to infinity.

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b) See the attached. This is what we did first. If you want to do this by hand, you find where the derivative is zero:

  6x^2 +6x -12 = 0

  6(x+2)(x-1) = 0 . . . . . local maximum at x=-2; local minimum at x=1.

You know the left-most zero of the derivative is the local maximum because of the nature of the curve (increasing, then decreasing, then increasing again).

The function values at those points are easily found by evaluating the function written in Horner form:

  ((2x +3)x -12)x

at x=-2, this is ((2(-2)+3)(-2) -12)(-2) = (2-12)(-2) = 20 . . . . (-2, 20)

at x = 1, this is 2 +3 -12 = -7 . . . . . . . . . . . . . . . . . . . . . . . . . (1, -7)

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c) The point of inflection of a cubic is the midpoint between the local extremes: ((-2, 20) +(1, -7))/2 = (-1, 13)/2 = (-0.5, 6.5)

A cubic curve is symmetrical about the point of inflection. When you consider the derivative is a parabola symmetric about the vertical line through its vertex, perhaps you can see why. The local extremes of the cubic are the zeros of the parabola, which are symmetric about that line of symmetry. Of course the vertex of the derivative (parabola) is the place where its slope is zero, hence the second derivative of the cubic is zero--the point of inflection.

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The cubic is concave down to the left of the point of inflection; concave up to the right of that point. The interval of downward concavity corresponds to the interval on which the first derivative (parabola) is decreasing or the second derivative is negative.