Question

1)Find the maximum and minimum values, if any, of the function given by
$f\left(x\right)=(2x-1{)}^2+3$

2)Find the maximum and minimum values, if any, of the function given by
$f\left(x\right)=9x^2+12x+2$

3)Find the maximum and minimum values, if any, of the function given by
$f\left(x\right)=-(x-1{)}^2+10$

4)Find the maximum and minimum values, if any, of the function given by
$f\left(x\right)=x^3+1$

Answer 1

1) Given,

$f\left(x\right)=(2x-1{)}^2+3$

Differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=4(2x-1)$

Putting $f^{\prime }\left(x\right)=0$

$4(2x-1)=0$

$\Rightarrow 2x-1=0$

$\Rightarrow x=\frac{1}{2}$

So, critical point is $x=\frac{1}{2}$

Using first derivative test,

$x<\frac{1}{2}\Rightarrow f^{\prime }\left(x\right)<0$

$x>\frac{1}{2}\Rightarrow f^{\prime }\left(x\right)>0$

Since, at $x=\frac{1}{2}$

sign of $f^{\prime }\left(x\right)$ changes from negative to positive,
So, it is a point of Minima.
Now,
$f\left(x\right)=(2x-1{)}^2+3$
Putting $x=\frac{1}{2}:$
$f\left(\frac{1}{2}\right)={(2\times \frac{1}{2}-1)}^2+3$
$=3$

Hence, minimum value of $f\left(x\right)$ is $3$ and there is no maximum value of $f\left(x\right)$.

2) $f\left(x\right)=9x^2+12x+2$

Differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=18x+12$

$\Rightarrow f^{\prime }\left(x\right)=6(3x+2)$

Putting $f^{\prime }\left(x\right)=0$

$6(3x+2)=0$

$\Rightarrow 3x+2=0$

$\Rightarrow x=-\frac{2}{3}$

So, critical point is $x=-\frac{2}{3}$

We know, $f^{\prime }\left(x\right)=18x+12$

Again, Differentiating w.r.t $x$

$f^{\prime \prime }\left(x\right)=18>0$

So, $x=-\frac{2}{3}$ is point of minima.
Now,
$f\left(x\right)=9x^2+12x+2$

Putting $x=-\frac{2}{3}$

$f\left(\frac{-2}{3}\right)=9{\left(\frac{-2}{3}\right)}^2+12\left(\frac{-2}{3}\right)+2$

$=4-8+2$

$=-2$

Hence, minimum value of $f\left(x\right)$ is $-2$ and there is no maximum value.

3) $f\left(x\right)=-(x-1{)}^2+10$

Differentiating w.r.t $x,$

$f^{\prime }\left(x\right)=-2(x-1)$

Putting $f^{\prime }\left(x\right)=0$

$\Rightarrow -2(x-1)=0$

$\Rightarrow (x-1)=0$

$\Rightarrow x=1$

So, critical point is $x=1$

we know, $f^{\prime }\left(x\right)=-2(x-1)$

Again, Differentiating w.r.t $x$

$f^{\prime \prime }\left(x\right)=-2<0$

So, $x=1$ is point of maxima

Now,
$f\left(x\right)=-(x-1{)}^2+10$
Putting $x=1$

$f\left(x\right)=-(1-1{)}^2+10$

$=10$

Hence, maximum value of $f\left(x\right)$ is $10$ and there is no minimum value of $f\left(x\right)$

4) $g\left(x\right)=x^3+1$

Differentiating w.r.t $x,$

$g^{\prime }\left(x\right)=3x^2$

Putting $g^{\prime }\left(x\right)=0$

$3x^2=0$

$\Rightarrow x=0$

So, critical point is $x=1$

$\because g^{\prime }\left(x\right)=3x^2$

$g^{\prime \prime }\left(x\right)=6x$

At $x=0$

$g^{\prime \prime }\left(0\right)=6\times 0=0$

$\therefore$ Point $x=0$ is neither a point of local maxima nor a point of local Minima, it is a point of inflexion.
Hence, there is no minimum or maximum value.