How do use the first derivative test to determine the local extrema f(x)=x^4-4x^3+4x^2+6 ?

Aug 8, 2015

You get: 2 minima and 1 maximum.

Explanation:

We can study the derivative of the function:
$f ' \left(x\right) = 4 {x}^{3} - 12 {x}^{2} + 8 x$
Set the derivative equal to zero:
$f ' \left(x\right) = 0$
$4 {x}^{3} - 12 {x}^{2} + 8 x = 0$
$4 x \left({x}^{2} - 3 x + 2\right) = 0$
With solutions:
${x}_{1} = 0$
${x}^{2} - 3 x + 2 = 0$ using the Quadratic Formula:
${x}_{2 , 3} = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2}$
two solutions:
${x}_{2} = 2$
${x}_{3} = 1$
We can now study the sign of the derivative setting: $f ' \left(x\right) > 0$ getting:

Graphically:
graph{x^4-4x^3+4x^2+6 [-20.27, 20.28, -10.14, 10.13]}