Question

I need help with problem 1 with a through explanation and solution please ​

Answer 1

Explanation:

The cubic ...

  f(x) = ax³ +bx² +cx +d

has derivatives ...

  f'(x) = 3ax² +2bx +c

  f''(x) = 6ax +2b

a)

By definition, there will be a point of inflection where the second derivative is zero (changes sign). The second derivative is a linear equation in x, so can only have one zero. Since it is given that a≠0, we are assured that the line described by f''(x) will cross the x-axis at ...

  f''(x) = 0 = 6ax +2b   ⇒   x = -b/(3a)

The single point of inflection is at x = -b/(3a).

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b)

The cubic will have a local extreme where the first derivative is zero and the second derivative is not zero. These will only occur when the discriminant of the first derivative quadratic is positive. Their location can be found by applying the quadratic formula to the first derivative.

  $x=\dfrac{-2b\pm\sqrt{(2b)^2-4(3a)(c)}}{2(3a)} = \dfrac{-2b\pm\sqrt{4b^2-12ac}}{6a}\\\\x=\dfrac{-b\pm\sqrt{b^2-3ac}}{3a}\qquad\text{extreme point locations when b^2>3ac }$

There will be zero or two local extremes. A local extreme cannot occur at the point of inflection, which is where the formula would tell you it is when there is only one.

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c)

Part A tells you the point of inflection is at x= -b/(3a).

Part B tells you the midpoint of the local extremes is x = -b/(3a). (This is half the sum of the x-values of the extreme points.) You will notice these are the same point.

The extreme points are located symmetrically about their midpoint, so are located symmetrically about the point of inflection.

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Additional comment

There are other interesting features of cubics with two local extremes. The points where the horizontal tangents meet the graph, together with the point of inflection, have equally-spaced x-coordinates. The point of inflection is the midpoint, both horizontally and vertically, between the local extreme points.