Question

For what values of x does the graph of `f(x)=x^3+3x^2+x+3` have a horizontal tangent?

Answer 1

`x = -3 +- sqrt(6)`

Explanation:

First, find the derivative.

`f'(x) = 3x^2 + 6x + 1`

A horizontal tangent will have a slope of `0`. The derivative represents the instantaneous rate of change of a function. Set the derivative to `0` and solve for `x`.

`0 = 3x^2 + 6x + 1`

`x= (-6 +- sqrt(6^2 - 4 * 3 * 1))/(2 * 3)`

`x = (-6 +- sqrt(24))/6`

`x = (-6 +- 2sqrt(6))/6`

`x = -1 +- 1/3sqrt(6)`

`x = -3 +- sqrt(6)`

Hopefully this helps!

Answer 2

`-1+-1/3sqrt 6`, The turning points can be located in the inserted Socratic graph.

Explanation:

`y=f(x)=x^3+3x^2+x+3`

`y'=3x^2+6x+1`=0, at x=`-1+-1/3 sqrt 6= -0.1835 and -1.816`, nearly.

So, the tangents at these points (-0.1835, 2.911) and (-1.815, 5.089) are

horizontal.

The equations to these tangents are

y = 2.911 and y = 5.089.

graph{(y-2.911)(y-5.089)(y-3-x-3x^2-x^3)=0 [-20, 20, -10, 10]}