Applications of Derivatives

Finding Critical Points

Find the critical points of f(x)=x³-3x by setting f'(x)=3x²-3=0.

Example

Set the derivative equal to zero to locate critical x-values.

highlighted = computed this step

Step 1 — Set up

Start with the function.

f(x)=x3−3xf(x)= x^{3}-3x

Step 2 — Differentiate

Differentiate to get f prime.

f′(x)=3x2−3f'(x)= \hlmath{3x^{2}-3}

Step 3 — Set derivative to zero

Set f prime equal to 0.

3x2−3=03x^{2}-3 = \hl{0}

Step 4 — Critical points

The critical points are x equals -1 and x equals 1.

x=-1x=1x= \hl{-1} \quad x= \hl{1}
critical-points A critical point occurs where f'(x)=0 or f'(x) is undefined. Setting the derivative equal to zero and solving gives the x-values where the function may have local maxima, minima, or saddle points.