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Use a graphing utility to graph the function to approximate (to 2 decimal places) any relative minima or maxima of the function. (If the answer does not exist, enter DNE)
h(x) = x^3 - 6x^2 + 15

relative minimum (x, y) = (______)
relative maximum (x, y)= (______)

Respuesta :

mhanifa

Given function:

  • h(x) = x³ - 6x² + 15

This is an odd degree function with positive leading coefficient. It means the graph is increasing and may have two turning points which are called local minimum/maximum or relative minimum/maximum.

As per graph it has:

  • Relative minimum at (4, - 17) and
  • Relative maximum at (0, 15).

The graph is attached.

Ver imagen mhanifa
semsee45

Answer:

Relative minimum (x, y) = (4, -17)

Relative maximum (x, y) = (0, 15)

Step-by-step explanation:

             

Given function:

[tex]h(x) = x^3 - 6x^2 + 15[/tex]

Use a graphing calculator to graph the function (see attachment).

The relative minima and maxima of a function are the turning points.

From inspection of the graphed function:

  • Relative minimum (x, y) = (4, -17)
  • Relative maximum (x, y) = (0, 15)

To find the x-values of the turning points, differentiate the function:

[tex]\implies h'(x)=3x^2-12x[/tex]

Then set the derivative of the function to zero and solve for x:

[tex]\implies 3x^2-12x=0[/tex]

[tex]\implies 3x(x-4)=0[/tex]

Therefore the x-values of the turning points are:

  • x = 0
  • x = 4

To find the y-values, substitute the x-values into the function:

[tex]\implies h(0)=(0)^3-6(0)^2+15=15[/tex]

[tex]\implies h(4)=(4)^3-6(4)^2+15=-17[/tex]

Therefore, this confirms that the maxima and minima are:

  • (0, 15) and (4, -17)
Ver imagen semsee45

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