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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Analytical

Critical Points and Extrema

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Critical Points and Extrema

Since our first derivative can tell us intervals of increasing/decreasing, it can also give us locations of Extrema.

Critical Points

A critical point of a function f(x)f(x) is a point xx ,in its domain, for which f′(x)=0\boxed{f'(x)=0}

Singular Points

A singular point of a function f(x)f(x) is a point xx ,in its domain, for which f′(x)f'(x) does not exist.


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Relative Extrema

A function f(x)f(x) has a relative (local) maximum at a point xx, in its domain, if there is a small interval II , around xx, such that f(x)≥f(y) f(x)\ge f(y)\ for every yy in II.



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A function f(x)f\left(x\right) has a relative (local) minimum at a point xx, in its domain, if there is a small interval II, around xx, such that f(x)≤f(y)f(x)\le f(y)for every yy in II.


Wize Tip
Relative extrema occur at either critical points or singular points!

Example: Critical Points and Extrema

Find the critical points and extrema of f(x)=2x3−3x2−36x+2f(x)=2x^{3}-3x^{2}-36x+2.

Taking the derivative of the given function, we get

f′(x)=6x2−6x−36=6(x2−x−6)=6(x−3)(x+2)\begin{aligned} f'(x)=&6x^2-6x-36\\ =&6(x^2-x-6)\\ =&6(x-3)(x+2)\\ \end{aligned}



Critical points: set f′(x)=0⇒x=−2, x=3f'(x)=0 \Rightarrow x=-2,~x=3


Setting f′(x)=0f'(x)=0 implies that x=−2,x=3x=-2,x=3 which are the two values for f(x)f(x). Now, consider the following table

From the table above, we can conclude that the given function f(x)f(x) is increasing on (−∞,−2)(-\infty,-2) and (3,∞)(3,\infty) and decreasing on (−2,3)(-2,3). Therefore f(x)f(x) has a relative maximum at x=−2x=-2 and a relative minimum at x=3x=3.

Conclusion: x=−2x=-2 is relative maximum, x=3x=3 is relative minimum.

Find the critical points for f(x)=x−3x3\displaystyle f(x)=x-3\sqrt[3]{x}