# "Determine where f(x)=3x^2 ln x has a horizontal tangent line?"

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The slope of a horizontal line is 0. The slope of a line tangent to the curve defined by y = f(x) at the point x = a is f'(a)

For `f(x) = 3x^2*ln x` ,

f'(x) = `3*x^2/x + 6x*ln x`

Toe determine where the tangent is a horizontal line solve `3*x^2/x + 6x*ln x = 0`

=> `3x + 6x*ln x = 0`

=> `3x(1 + 2*ln x) = 0`

x = 0

and `1 + 2*ln x = 0`

=> `x = 1/sqrt e` and `x= -1/sqrt e`

But for x = 0 and `x = -1/sqrt e` the logarithm is not defined.

**The tangent line to `f(x) = 3x^2*ln x` is horizontal only where **`x = 1/sqrt e`