Determine the exact values of the global extrema
of these functions \(f\) on their indicated domains.
These exercises were designed to be most beneficially done with only pen-and-paper.
Answers can be verified visually using a graphing calculator though.
The phrase “for \(x \in [a,b]\)” just indicates the domain of \(f.\)
\( \displaystyle f(x) = 2(x-3)^2+5 \\[1ex]\quad\text{ for } x \in [0,4]\)
\( \displaystyle f(x) = x^2-5x+6 \\[1ex]\quad\text{ for } x \in [0,4]\)
\( \displaystyle f(x) = \tfrac{1}{3}x^3-4x^2+7x \\[1ex]\quad\text{ for } x \in [0,10]\)
\( \displaystyle f(x) = 12+45x+3x^2-x^3 \\[1ex]\quad\text{ for } x \in [-6,6]\)
\( \displaystyle f(x) = x^9-3x^2+4 \\[1ex]\quad\text{ for } x \in [0,2]\)
\( \displaystyle f(x) = x^5+\tfrac{7}{x^3}-5 \\[1ex]\quad\text{ for } x \in [1,2]\)
\( \displaystyle f(x) = 47-31x^4+7x^7 \\[1ex]\quad\text{ for } x \in [-1,2]\)
\( \displaystyle f(x) = x^3-3x^2-6x+3 \\[1ex]\quad\text{ for } x \in [-2,5]\)
\( \displaystyle f(x) = 40+12x-3x^2-x^3 \\[1ex]\quad\text{ for } x \in [-4,2]\)
\( \displaystyle f(x) = \tfrac{1}{3}x^3-3x^2+7x+1 \\[1ex]\quad\text{ for } x \in [0,5]\)
\( \displaystyle f(x) = x^3-2x^2-x+2 \\[1ex]\quad\text{ for } x \in [-1,3]\)
\( \displaystyle f(x) = \tfrac{x^2+15x+54}{2x-1} \\[1ex]\quad\text{ for } x \in (0,16]\)
\( \displaystyle f(x) = \tfrac{4x-12}{x^2-4x+4} \\[1ex]\quad\text{ for } x \in (-\infty, \infty)\)
\( \displaystyle f(x) = \tfrac{1}{x^3-4x^2+x+6} \\[1ex]\quad\text{ for } x \in (-1,3)\)
\( \displaystyle f(x) = 2\sin(x)-x+8 \\[1ex]\quad\text{ for } x \in \bigl[0, 2\pi\bigr]\)
\( \displaystyle f(x) = 2\sin(x)-\sin(2x) \\[1ex]\quad\text{ for } x \in \bigl[0, 2\pi\bigr]\)
\( \displaystyle f(t) = \mathrm{e}^{\tfrac{1}{5}t} + \mathrm{e}^{-\tfrac{1}{2}t} \\[1ex]\quad\text{ for } t \in (-\infty, \infty)\)
\( \displaystyle f(x) = x\mathrm{e}^x \\[1ex]\quad\text{ for } x \in [-3,3]\)
\( \displaystyle f(x) = \tfrac{1}{\mathrm{e}^x + \mathrm{e}^{-2x}} \\[1ex]\quad\text{ for } x \in [0,\infty)\)
\( \displaystyle f(x) = \ln\bigl(1-x+x^2\bigr) \\[1ex]\quad\text{ for } x \in [0,3]\)
\( \displaystyle f(x) = \ln\bigl(\cos(x)\bigr)+x \\[1ex]\quad\text{ for } x \in \bigl(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr)\)
A cubic polynomial of the form
\(\tfrac{1}{3}x^3 - \tfrac{a+b}{2}x^2 + (ab)x + C\)
will have extrema at \(x=a\) and \(x=b.\)
More generally a cubic polynomial of the form
\(\tfrac{1}{3}x^3 - hx^2 + (h^2-d)x + C\)
will have extrema at \(x = h+\sqrt{d}\) and \(x = h-\sqrt{d}.\)
For a cubic polynomial with roots \(r_1\) and \(r_2\) and \(r_3,\)
its extrema will be located at \(x = \sum r_i \pm \sqrt{\sum r_ir_j}
(the first sum has three summands, and the radicand sum has six).
For \(m \gt n\) a polynomial of the form \(x^m + ax^n + C\)
will have an extrema at zero and at \(x = \pm \sqrt[m-n]{-\tfrac{an}{m}};\)
in particular, one of \(m\) or \(n\) or \(a\) ought to be negative,
and there will be one or two extrema besides zero
depending on the parity of \(m-n.\)
For \(m \gt n \gt k\) a polynomial of the form \(x^m + ax^n + bx^k + C\)
will have a derivative with a factor of quadratic form,
\(\square^2 + \square + \square,\) only if \(m = 2n-k,\)
and will have real extrema only if \((an)^2 - 4mbk \gt 0.\)
Both of the functions \(\mathrm{e}^{ax} + \mathrm{e}^{-bx}\)
and \(\tfrac{1}{\mathrm{e}^{ax} + \mathrm{e}^{-bx}}\)
have a single extremum at \(x = \tfrac{1}{a+b}\log_a(b),\)
notably positive for \(a \lt b\) and a tiny number
unless \(a\) and \(b\) are tiny themselves.
Any rational function \(P/Q\) will have calculable extrema
so long as \(\operatorname{deg}P + \operatorname{deg}Q \leq 3.\)
The rational function \(\tfrac{(x-a)(x-b)}{x-c}\) for \(c \gt a \gt b\)
will have extrema at \(c \pm \sqrt{(c-a)(c-b)}.\)