Answer to Question #213700 in Calculus for anuj

Question

a) find and classify the critical points of the functions f(x) = 2x^3 + 3x^2 - 12 x +1 into maximum, minimum and inflection points as appreciate.

(b) The sum of two positive numbers is S. find the maximum value of their product.

Answer 1

$i)\ f(x)=2x^{3}+3x^{2}-12x+1$

For this $f(x)$ , we have to find critical points , maximum and minimum points and inflections also of this given functions .

$f^{'}(x)=6x^{2}$ $+\ 6x-12$

To find the critical values , we differentiate and find values of $x,$ that is we put $f^{'}(x)=0$ ,

$f^{'}(x)<0\implies$ $f(x)$ is decreasing

$f^{'}(x)=0\implies$ $f(x)$ is stationary

$f^{'}(x)>0\implies$ $f(x)$ is increasing.

$f^{'}(x)=6x^{2}+6x-12$

$f{'}(x)=0$

$\implies$ $6x^{2}+6x-12=0$

At a critical point , $f{'}(x)=0$

$\implies$ $x^{2}+x-2=0$

${\therefore}$ $x=-2,1$

$x=-2\implies$ $f(-2)=20$

$x=1\implies$ $f(1)=-7$

Differentiating , with respect to $x,$

$f^{''}(x)=12x+6$

$x=-2\implies f^{''}(-2)=$ $-8<0$ , maximum

$x=1\implies\ f^{''}(1)=18>0$ , minimum

Inflection points , for the function -

$f(x)=2x^{3}+3x^{2}-12x+1$

An inflection point is a point on a curve where concavity changes from concave up to concave down or vice versa .

$f^{'}(x)=6x^{2}+6x-12$

$f^{''}(x)=12x+6$

Inflection point also has same meaning as where $f^{''}(x)$ switches sign -

$=12x+6=0$

$=x=\dfrac{-1}{2}$

Putting the obtained $x$ in the equation $f(x) ,$ we get -

$y=\dfrac{15}{2}$

So point of inflection are -$(\dfrac{-1}{2},\dfrac{15}{2})$

$ii)$ Let the two numbers be $x\ and \ y.$ .

$=x+y=S$ .................................1)

$=$ $y=S-x$

their product is given as -

$=xy=x(S-x)=Sx-x^{2}$

$f(x)=Sx-x^{2}$

$f^{'}(x)=S-$ $2x$

putting $f^{'}(x)=0$ , we get -

$x=\dfrac{S}{2}$

Putting the value of $x\$ in equation 1) , we get ,

$y=\dfrac{S}{2}$

The maximum value of product can be written as -

$xy$ , Putting the value of $x\ and \ y-$

$=\dfrac{S}{2}\times\dfrac{S}{2}=\dfrac{S^{2}}{4}$

Which is required product