Question

Find the local maxima and local minima, if any of the following functions. Also, find the local maximum and the local minimum values, as the case may be as follows.
$f\left(x\right)=x^2$

$g\left(x\right)=x^3-3x$

$h\left(x\right)=sinx+cosx,0<x<\frac{\pi }{2}$

$f\left(x\right)=sinx-cosx,0<x<2\pi$

$f\left(x\right)=x^3-6x^2+9x+15$

$g\left(x\right)=\frac{x}{2}+\frac{2}{x},x>0$

$g\left(x\right)=\frac{1}{x^2+2}$

$f\left(x\right)=x\sqrt{1-x},x>0$

Answer 1

Given function is, $f\left(x\right)=x^2$
$\therefore$ Differentiating w.r.t. x, f'(x)=2x~ and ~again differentiating w.r.t x, we get f"(x)=2
for maxima or minima, put f'(x)=0
$\therefore 2x=0\Rightarrow x=0$
Thus, x=0 is the only critical point which could be possibly the point of local maxima or local minima of f.
We have $f"\left(0\right)=2>0$ (which is positive)
Therefore, by second derivative test, x=0 is a point of local minima and local minimum value of f at x=0 is $f\left(0\right)=x^2=0^2=0$

Given function is, $g\left(x\right)=x^3-3x$
$\therefore g^{\prime }\left(x\right)=3x^2-3and{g}^{\prime \prime }\left(x\right)=6x$
For maxima or minima put g'(x)=0
$\therefore 3x^2-3=0\Rightarrow x=\pm 1$
Thus, we expect extremum only at two points -1 and 1
At $x=-1,g"(-1)=6(-1)=-6<0$
$\therefore$ g has a local maxima at x=-1 and local maximum value = g(-1)
$=(-1{)}^3-3(-1)=-1+3=2$
At $x=1,g"\left(1\right)=6\times 1=6>0$
$\therefore$ g has a local minima at x=1 and local minimum value =g(1)
$=(1{)}^3-3(1)=1-3=-2$
At $x=1,g"\left(1\right)=6\times 1=6>0$
Therefore g has a local minima at x=1 and local minimum value = g(1)
$=1^3-3\times 1=-2$

Given function is, $h\left(x\right)=sinx+cosx,0<x<\frac{\pi }{2}$
$\therefore h^{\prime }\left(x\right)=cosx-sinx$ and $h"\left(x\right)=-sinx-cosx$
For maxima or minima put h'(x)=0
$\Rightarrow sinx=cosx\Rightarrow \frac{sinx}{cosx}=1\Rightarrow tanx=1\Rightarrow x=\frac{\pi }{4}ϵ(0,\frac{\pi }{2})$
At $x=\frac{\pi }{4},h"\left(\frac{\pi }{4}\right)=-sin\frac{\pi }{4}-cos\frac{\pi }{4}=-sin\frac{\pi }{4}-cos\frac{\pi }{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$
Therefore, by second derivative test $x=\frac{\pi }{4}$ is a point of local maxima and the local maximum value of h at $x=\frac{\pi }{4}$ is
$h\left(\frac{\pi }{4}\right)=sin\frac{\pi }{4}+cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Given function is, $f\left(x\right)=sinx-cosx,0<x<2\pi$
$\therefore f^{\prime }\left(x\right)=cosx+sinxandf"\left(x\right)=sinx+cosx$
For maxima put f'(x)=0
$\Rightarrow cosx+sinx=0\Rightarrow \frac{sinx}{cosx}=-1,i.e.tanx=-1\Rightarrow x=\pi -\frac{\pi }{4},2\pi -\frac{\pi }{4}\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4},xϵ(2,2\pi )$
$\therefore$ The points at which extremum may occur are $\frac{3\pi }{4}$ and $\frac{7\pi }{4}$
At $x=\frac{\pi }{4},f"\left(\frac{3\pi }{4}\right)=-sin\frac{3\pi }{4}+cos\frac{3\pi }{4}=-sin(\pi -\frac{\pi }{4})+cos(\pi -\frac{\pi }{4})$
$[\because sin(\pi -\theta )=sin\theta andcos(\pi -\theta )=-cos\theta ]$
$=-sin\frac{\pi }{4}-cos\frac{\pi }{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$
$\therefore x=\frac{3\pi }{4}$ is a point of maxima.
Maxima value $=f\left(\frac{3\pi }{4}\right)=sin\frac{3\pi }{4}-cos\frac{3\pi }{4}$
$=sin(\pi -\frac{\pi }{4})-cos(\pi -\frac{\pi }{4})=sin\frac{\pi }{4}-cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
At $x=\frac{7\pi }{4},f"\left(\frac{7\pi }{4}\right)+cos\frac{7\pi }{4}=-sin(2\pi -\frac{\pi }{4})+cos(2\pi -\frac{\pi }{4})$
$[\because sin(2\pi -\theta )=-sin\theta andcos(2\pi -\theta )=cos\theta ]=sin\frac{\pi }{4}+cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$
$\therefore x=\frac{7\pi }{4}$ is a point of minima.
Minimum value
$f\left(\frac{7\pi }{4}\right)=sin\frac{7\pi }{4}-cos\frac{7\pi }{4}=sin(2\pi -\frac{\pi }{4})-cos(2\pi -\frac{\pi }{4})=-sin\frac{\pi }{4}-cos\frac{\pi }{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}$

Given functions is $f\left(x\right)=x^3-6x^2+9x+15$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-12x+9andf"\left(x\right)=6x-12$
For maxima or minima put f'(x)=0
$\Rightarrow 3x^2-12x+9=0\Rightarrow 3(x^2-4x+3)=0\Rightarrow 3(x^2-3x-x+3)=0\Rightarrow 3\left[x\right(x-3)-1(x-3\left)\right]=0\Rightarrow 3(x-1)(x-3=0\Rightarrow x=1orx=3$
At $x=1f"\left(1\right)=6\times 1-12=6-12=-6<0$
$\therefore$ x=1 is a point of maxima.
Maximum value $=f\left(1\right)=(1{)}^3-6(1{)}^2+9\left(1\right)+15=1-6+9+15=19$
At $x=3,f"\left(3\right)=6\times 3-12=18-12=6>0$
x=3 is a point of minima
$\therefore$ Minimum value $=f\left(3\right)=3^3-6\times 3^2+9\times 3+15=27-54+27+15=15$

Given function is $g\left(x\right)=\frac{x}{2}+\frac{2}{x},x>0$
$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{2}+2(-\frac{1}{x^2})=\frac{1}{2}-\frac{2}{x^2}andg"\left(x\right)=-2(-2)x^{-3}=\frac{4}{x^3}$
For maxima or minima put, g'(x)=0
$\Rightarrow \frac{1}{2}-\frac{2}{x^2}=0\Rightarrow \frac{1}{2}=\frac{2}{x^2}{\Rightarrow }^2=4\Rightarrow x=2,-2$ [$\because x>0$ (given)]
At x=2, $g"\left(2\right)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}>0$
$\therefore x=2$ is a point of minima.
Minimum value, $g\left(2\right)=\frac{2}{2}+\frac{2}{2}=1+1=2$

Given function is $g\left(x\right)=\frac{1}{x^2+2}$
Now, $g\left(x\right)=(x^2+2{)}^{-1}\Rightarrow g^{\prime }(x)=-(x^2+2{)}^{-1-1}\times 2x=\frac{-2x}{(x^2+2{)}^2}g"\left(x\right)=\frac{(x^2+2{)}^2.(-2)-(-2x\left).2\right(x^2+2).2x}{(x^2+2{)}^4}=\frac{-2(x^2+2{)}^2+8x^2(x^2+2)}{(x^2+2{)}^4}=\frac{(x^2+2)[-2x^2-4+8x^2]}{(x^2+2{)}^4}=\frac{6x^2-4}{(x^2+2{)}^3}=\frac{2(3x^2-2)}{(x^2+2{)}^3}$
For maxima or minima put $g^{\prime }\left(x\right)=0\Rightarrow \frac{-2x}{(x^2+2{)}^2}=0\Rightarrow -2x=0\Rightarrow x=0$
At x=0, $g"\left(0\right)=\frac{2\left[3\right(0{)}^2-2]}{\left[\right(0{)}^2+2{]}^3}=\frac{-4}{8}=-\frac{1}{2}<0$
$\therefore x=0$ is a point of maxima. Maximum value, $g\left(0\right)=\frac{1}{(0{)}^2+2}=\frac{1}{2}$

Given function is $f\left(x\right)=x\sqrt{1-x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x(-1)}{2\sqrt{1-x}}+\sqrt{1-x}=\frac{-x+2(1-x)}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}f"\left(x\right)=\frac{1}{2}\left[\frac{\sqrt{1-x}(-3)-(2-3x)\left(\frac{-1}{2\sqrt{1-x}}\right)}{(1-x)}\right]=\frac{\sqrt{1-x}(-3)-(2-3x)\left(\frac{-1}{2\sqrt{1-x}}\right)}{2(1-x)}=\frac{-6(1-x)+(2-3x)}{4(1-x{)}^{\frac{3}{2}}}=\frac{3x-4}{4(1-x{)}^{\frac{3}{2}}}$
For maxima or minima put f'(x)=0
$\frac{2-3x}{2\sqrt{1-x}}=0\Rightarrow 2-3x=0\Rightarrow x=\frac{2}{3}Atx=\frac{2}{3},f"\left(\frac{2}{3}\right)=\frac{3\left(\frac{2}{3}\right)-4}{4{(1-\frac{2}{3})}^{\frac{3}{2}}}=\frac{2-4}{4{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}=\frac{-1}{2{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}<0$
$\therefore x=\frac{2}{3}$ is a point of maxima.
Maximum value $=f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3}\times \sqrt{\frac{1}{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}$