How do you find the points where the graph of the function f (x) = 3 x^ 3 + 8 x^ 2 + 4 x + 7 has horizontal tangents and what is the equation?

1 Answer
Jul 19, 2016

Horizontal tangent equations:
color(white)("XXX")color(green)(f(x)=6.438995) at color(green)(""(-0.30094,6.438995))
color(white)("XXX")color(green)(f(x)=8.877877) at color(green)(""(-1.47683,8.877877))

Explanation:

The slope of the tangent is given by the derivative of the function.

Given
color(white)("XXX")f(x)=3x^3+8x^2+4x+7
the slope of the tangent for a general value of x is
color(white)("XXX")f'(x)=9x^2+16x+4

The tangent is horizontal when the slope is equal to zero.

color(white)("XXX")9x^2+16x+4=0 for values of x where the tangent is horizontal

Using the quadratic formula, we have
color(white)("XXX")x=(-16+-sqrt(16^2+4xx9xx4))/(2xxx9)=(-8+-2sqrt(7))/9

Using a calculator, we get (approximate) values
color(white)("XXX")x=-0.30094 and x=-1.47683

Plugging each of these values for x into the original equation
gives corresponding values for f(x)
color(white)("XXX")f(x)=6.438995 and f(x)=8.877877
which are the equations of the corresponding horizontal lines.

Here is a graph of the original function and the tangent lines derived above which supports this answer:
enter image source here