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What is the maximum number of turns in the graph of this function?
f(x)=4x³-12x2-x+15
Answer here


Sagot :

Answer:

 2

Step-by-step explanation:

You want the maximum number of turning points in the graph of the cubic function ...

  f(x) = 4x³ -12x² -x +15

Turning points

In general, the maximum number of turning points in the graph of a polynomial function is 1 less than its degree. The given function has degree 3, so the maximum number of turning points is 3 - 1 = 2.

  f(x) has 2 turning points (maximum).

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Additional comment

The turning points in the graph are found where the first derivative of the function is zero. The first derivative of f(x) is ...

  f'(x) = 12x² -24x -1

The discriminant of this quadratic is ...

  b² -4ac = (-24)² -4(12)(-1) = 576 +48 = 624

This value is positive, indicating the quadratic has two (2) real zeros. There is a turning point at each of those.

If the first derivative has one zero, that will be a point of inflection in the graph that will look like a "flat spot." The tangent is horizontal there, but the graph does not actually change the sign of its slope.

When the first derivative of a cubic function has no real zeros, there are no turning points in its graph. The sign of the slope of the graph will be the sign of the function's leading coefficient. The slope will vary, but will not change sign.

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