Second-derivative test
\[f'(c)=0:\quad f''(c)\lt 0\Rightarrow\text{local max},\qquad f''(c)\gt 0\Rightarrow\text{local min}\]
Applications of Derivatives
\(f\) has a local maximum at \(c\) if \(f(c)\ge f(x)\) for all \(x\) near \(c\), and a local minimum if \(f(c)\le f(x)\) near \(c\). At such a point (if \(f\) is differentiable) \(f\,'(c)=0\): \(c\) is a critical point.
| Test | Local maximum | Local minimum |
|---|---|---|
| First derivative | \(f\,'\) changes from + to − | \(f\,'\) changes from − to + |
| Second derivative | \(f\,'(c)=0,\ f\,''(c)<0\) | \(f\,'(c)=0,\ f\,''(c)>0\) |
If \(f\,''(c)=0\) the second-derivative test fails; use the first-derivative test. A critical point where \(f\,'\) does not change sign is a point of inflection (e.g. \(x^3\) at 0).
A continuous function on a closed interval attains an absolute maximum and minimum. Evaluate f at every critical point in (a, b) and at a and b; the largest value is the absolute maximum, the smallest the absolute minimum.
\[f'(c)=0:\quad f''(c)\lt 0\Rightarrow\text{local max},\qquad f''(c)\gt 0\Rightarrow\text{local min}\]
Compare the values at critical points inside the interval and at the two end points.
The absolute maximum of \(x^3-3x\) on \([0,3]\) is at the end point \(x=3\) (value 18), not at a critical point.
A local maximum need not be the largest value overall; a local minimum value can even exceed a local maximum value.
Squares of side \(x\) are cut from the corners of a 30 cm × 30 cm sheet and the sides folded up.
\(V=x(30-2x)^2\), \(V'=(30-2x)(30-6x)=0\Rightarrow x=5\) (\(x=15\) gives \(V=0\)). \(V_{\max}=5\cdot20^2=2000\) cm³.
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Start PracticeFind critical points, classify them, and for closed intervals compare with the end points. In word problems, reduce to one variable using the constraint.
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