Question

Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. $$ y=-x^{4}+3 x^{2}-1 $$

Short answer

The graph of the function has horizontal tangents at points (0,-1), \(\left(\sqrt{\frac{3}{2}}, -\frac{1}{2}\right)\), and \(\left(-\sqrt{\frac{3}{2}}, -\frac{1}{2}\right)\).

Step-by-step solution

Find the derivative

The derivative of the function \(y = -x^4 + 3x^2 - 1\) can be found via the power rule of differentiation, which states that the derivative of \(x^n\) is \(n*x^{n-1}\). So, \(y' = -4x^3 + 6x\).

Set the derivative equal to zero

To find where the function's tangent line is horizontal, set the derivative equal to zero and solve for x. \(-4x^3 + 6x = 0\). Factor out \(x\) to get \(x(-4x^2 + 6) = 0\). This gives possible x-values as \(x = 0, \sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}}\).

Find the y-values

Substitute \(x = 0, \sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}}\) into the original function \(y = -x^4 + 3x^2 - 1\) to find the corresponding y-values. This results in points (0,-1), \(\left(\sqrt{\frac{3}{2}}, -\frac{1}{2}\right)\), \(\left(-\sqrt{\frac{3}{2}}, -\frac{1}{2}\right)\).