For what values of x does the graph of f(x)=x³+3x²+x+3 have a horizontal tangent?

Solution:

Given f(x)=x³+3x²+x+3

Equation of tangent can be given as a derivative of f(x).

f'(x) = 3x²+6x+1

The derivative of f(x) or the slope of the tangent = 0 as it is a horizontal line.

Thus 3x²+6x+1 = 0

Using quadratic formula, x= -b±√(b² -4ac) / 2a

x= \(\dfrac{-3\pm \sqrt{6}}{3}\)

x = -0.183, -1.816

So, the required values of x are -0.183, -1.816

For what values of x does the graph of f(x)=x³+3x²+x+3 have a horizontal tangent?

Summary:

The values of x for which the graph of f(x)=x³+3x²+x+3 has a horizontal tangent are -0.183, -1.816