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

Aug 23, 2015

The local extrema occur at point where the first derivative is equal to zero (critical points) and where the slope (as given by the first derivative) switches between positive and negative

Explanation:

Given $f \left(x\right) = 4 {x}^{3} - 3 {x}^{4}$

$\frac{d \left(f \left(x\right)\right)}{\mathrm{dx}} = 12 {x}^{2} - 12 {x}^{3}$

For critical points
$\textcolor{w h i t e}{\text{XXXX}} 12 {x}^{2} - 12 {x}^{3} = 0$

$\rightarrow \textcolor{w h i t e}{\text{XX}} x \cdot x \cdot 12 \cdot \left(1 - x\right) = 0$

$\rightarrow \textcolor{w h i t e}{\text{XX}} x = 0 \mathmr{and} x = 1$ are critical points

Consider three points:
$\textcolor{w h i t e}{\text{XXX}} x = - 1$ (below the smallest critical point)
$\textcolor{w h i t e}{\text{XXX}} x = \frac{1}{2}$ (between the critical points)
$\textcolor{w h i t e}{\text{XXX}} x = 2$ (above the largest critical point)

$\frac{\mathrm{df}}{\mathrm{dx}} \left(- 1\right) = 12 \cdot {\left(- 1\right)}^{2} - 12 \cdot {\left(- 1\right)}^{3} = 24$

$\frac{\mathrm{df}}{\mathrm{dx}} \left(\frac{1}{2}\right) = 12 \cdot {\left(\frac{1}{2}\right)}^{2} - 12 \cdot {\left(\frac{1}{2}\right)}^{3} = \frac{12}{8}$

$\frac{\mathrm{df}}{\mathrm{dx}} \left(2\right) = 12 \cdot {\left(2\right)}^{2} - 12 \cdot {\left(2\right)}^{3} = - 48$

The slope remains positive on each side of $x = 0$
so $x = 0$ gives a point of inflection which is not a local extrema.

The slope changes from positive to negative on the two sides of $x = 1$
so $f \left(1\right) = 1$ is a local maximum.

The situation can be further examined by considering the graph below:
graph{4x^3-3x^4 [-2.733, 2.744, -1.366, 1.374]}