Mathematics: Find the equations of any horizontal tangents to the curve y = x^3 - 3x ^2 - 9x + 2

Find the equations of any horizontal tangents

donnasueanger

04.02.2022

Answers:

y = -25

y = 7

Step-by-step explanation:

Find the derivative of the function.

$\frac{dy}{dx} y = 3x^2-6x-9$

The zeros of the derivative are the local extrema (Maximum & Minimum)

0 = 3x^2-6x-9

Take out the greatest common factor, 3, to get this equation:

0 = 3(x^2-2x-3)

Factor out the brackets.

0 = 3(x - 3) ( x + 1 )

You are left with two possible values for x:

x = 3

x = -1

By substituting each value into the original function y = x^3 - 3x^2 - 9x + 2 , you get the equation of any horizontal tangents to the curve.

When x = 3:

y = (3)^3 - 3(3)^2 - 9(3) + 2 = -25 Meaning y = -25

When x = -1:

y = (-1)^3 - 3(-1)^2 - 9(-1) + 2 = 7 Meaning y = 7

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dee5896

01.12.2020

Option 2

Step-by-step explanation:

56c¹⁶/162c³

28c¹³/81

Sqrt(28c¹³/81)

Sqrt(2²×7×c¹²×c/9²)

(2c⁶/9)sqrt(7c)

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