Maximum and Minimum of a Function

Assume that \(f\) attains a local maximum at \(a\). There is an interval \((p,q)\) such that \(a\in(p,q)\) such that \(f(x)\leq f(a)\) for all \(x\in(p,q)\). Since \(f\) is differentiable at \(a\) we have that \[f^{\prime}(a)=\lim_{\varepsilon\to 0}\frac{f(a+\varepsilon)-f(a)}{\varepsilon}.\] If \(|\varepsilon|\) is small enough such that \(a+\varepsilon\in (p,q)\) we must have \(f(a+\varepsilon)\leq f(a)\). For \(\varepsilon > 0\) we get \(\frac{f(a+\varepsilon)-f(a)}{\varepsilon}\leq 0\), and for \(\varepsilon < 0\) we have \(\frac{f(a+\varepsilon)-f(a)}{\varepsilon}\geq 0\). Hence: \[\lim_{\varepsilon\to 0+}\frac{f(a+\varepsilon)-f(a)}{\varepsilon}\leq 0\quad\quad\quad\quad\quad\mbox{ and }\quad\quad\quad\quad\quad \lim_{\varepsilon\to 0-}\frac{f(a+\varepsilon)-f(a)}{\varepsilon}\geq 0.\] Since the last two limits have to be the same and equal to \(f^{\prime}(a)\) we conclude that \(f^{\prime}(a)=0\).

Critical values are those numbers \(c\) such that \(f^{\prime}(c)=0\) or \(f^{\prime}\) does not exist. Since \(f\) is differentiable for every \(x\), all critical values must satisfy \(f^{\prime}(c)=0\).

We have \(f^{\prime}(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2)\). Therefore the critical values are \(x=-2\) and \(x=3\).

Critical values are those numbers \(c\) such that \(f^{\prime}(c)=0\), \(f^{\prime}\) does not exist, or points on the boundary. Since \(f\) is differentiable for every \(x\), all critical values must satisfy \(f^{\prime}(c)=0\).

We have \(f^{\prime}(x)=1+2\cos x\). The equation \(\cos x=-\frac12\) has two solutions in \(\left[\frac{\pi}2,200\right]\). The solutions are \(x=\frac{2\pi}3\) and \(x=\frac{4\pi}3\). Hence the critical values are \(\frac{\pi}2\), \(\frac{2\pi}3\), \(\frac{4\pi}3\), and \(200\). The values of \(f\) at these points are: \(f\left(\frac{\pi}2\right)=\frac{\pi}2+2\), \(f\left(\frac{2\pi}3\right)=\frac{2\pi}3+\sqrt 3\), \(f\left(\frac{4\pi}3\right)=\frac{4\pi}3-\sqrt 3\), and \(f(200)=200+2\sin(200)\). Clearly, \(f(200)\) is the largest of the four numbers since \(\sin x\geq -1\) implies \(f(200)\geq 200+2\cdot (-1)=198\). Therefore the maximum is attained for \(x=200\) and is equal to \(200+2\sin(200)\).

Let us prove the inequality \(f\left(\frac{2\pi}3\right) > f\left(\frac{\pi}2\right)\), i.e. \[\frac{2\pi}3+\sqrt 3 > \frac{\pi}2+ 2.\] This inequality is equivalent to \(4\pi+6\sqrt 3 > 3\pi+12\), which in turn is equivalent to \(\pi+6\sqrt 3 > 12\). The last inequality is true because \(\pi > 3\) and \(3 > 1.7\) hence \(\pi+6\sqrt 3 > 3+6\cdot 1.7=3+10.2=13.2 > 12\). This implies that \(f\left(\frac{2\pi}3\right) > f\left(\frac{\pi}2\right)\).

We will now prove that \(f\left(\frac{4\pi}3\right) < f\left(\frac{\pi}2\right)\), i.e. \[ \frac{4\pi}3-\sqrt 3 < \frac{\pi}2+2. \] After multiplying by 6, the last inequality becomes equivalent to \(8\pi -6\sqrt 3 < 3\pi+12\), which is equivalent to \(5\pi < 12+6\sqrt 3\). The last inequality is true because \(12+6\sqrt 3 > 12+6\cdot 1.7 > 12+10=22 > 5\cdot 4 > 5\cdot \pi\).

The minimum is attained for \(x=\frac{4\pi}3\) and the maximum is attained for \(x=200\).

Notice that \(f^{\prime}(x)= x^3-x^2-4x+4= (x-2)(x+2)(x-1)\). Hence the critical values are the

• Endpoints \(-3\) and \(3\);
• Points for which the derivative is \(0\), i.e. \(-2\), \(1\), and \(2\);
• Points at which the derivative is not defined (there are no such points).

We now calculate: \[f(-3)=\frac{81}{4}+\frac{27}3-2\cdot 9-4\cdot 3+5=\frac{17}4,\] \[f(-2)=4+\frac83-8-8+5=-\frac{13}3,\] \[f(1)=\frac14-\frac13-2+4+5=\frac{3-4+84}{12}=\frac{83}{12},\] \[f(2)=4-\frac83-8+8+5=\frac{19}3,\] \[f(3)=\frac{81}4-\frac{27}3-2\cdot 9+4\cdot 3+5=\frac{41}4.\]

Hence the absolute maximum is \(\frac{83}{12}\) and is attained at \(x=1\). The absolute minimum is \(-\frac{13}3\) and it is attained for \(x=-2\).

We first take the derivative \(f^{\prime}(x)=\frac17x^{-\frac67}\cdot (2-x)+x^{\frac17}\cdot (-1)=\frac1{x^{\frac67}}\cdot \left(\frac17(2-x)-x\right)=\frac{ 2-8x}{7x^{\frac67}}\). The critical values are the points \(a\) for which \(f^{\prime}(a)=0\), or the function \(f\) is not differentiable at \(a\). The given function is not differentiable for \(a=0\), and \(f^{\prime}\left(\frac14\right)=0\). Hence the critical values are \(0\) and \(\frac14\).

Critical values are those numbers \(c\) such that \(f^{\prime}(c)=0\) or \(f^{\prime}\) does not exist. Since \(f\) is not differentiable at \(x=5\), the number \(x=5\) is a critical value.

For \(x > 5\) we have \(f(x)=x^2-x+2\) hence \(f^{\prime}(x)=2x-1\). The solution to the equation \(f^{\prime}(x)=0\) does not satisfy \(x > 5\), hence there are no critical values bigger than \(5\).

For \(x < 5\) we have \(f(x)=x^2-3x+12\). We now have \(f^{\prime}(x)=2x-3\) and the solution to the equation \(f^{\prime}(x)=0\) is \(x=\frac32\). Since \(\frac32 < 5\), this is a critical value.