Question

Let $f(x)={{x}^{3}}+b{{x}^{2}}+cx+d$, \(0 < {b^2} < c\). Then $f$
1) is bounded
2) has a local maxima
3) has a local minima
4) is strictly increasing

Answer 1

Hint:To solve this question we should know the concept of differentiation. To solve this question we will have to differentiate the function given to us and see the conditions we will get after differentiation. If ${f}'(x)$ is greater than zero then the function will be strictly increasing and if it will be equal to zero then the function will either have local maxima or local minima.

Complete step-by-step solution:
The question ask us to find whether the function is bounded or has a local minima or is strictly increasing or has a local Maxima when a function given is $f(x)={{x}^{3}}+b{{x}^{2}}+cx+d$ and condition applied is \(0 < {b^2} < c\). The function given to us is:
$\Rightarrow f(x)={{x}^{3}}+b{{x}^{2}}+cx+d$
So now we will be differentiating the function given above. The formula being applied is if a function $g(x)={{x}^{n}}$ is given then the differentiation of the function becomes${g}'\left( x \right)=n{{x}^{n-1}}$ . Applying the same in the expression given to us in the question we will get:
$\Rightarrow {f}'(x)=3{{x}^{2}}+2bx+c$
Expression we got after differentiation is a quadratic equation. If the differentiation of the function, ${f}'(x)$ is greater than zero this shows that the function is strictly increasing. To check the ${f}'(x)$, we will find the discriminant of the function. So discriminant of the function will be:
$\Rightarrow D=4{{b}^{2}}-4\times 3\times c$
$\Rightarrow D=4\left( {{b}^{2}}-3\times c \right)$
The condition given in the question is \(0 < {b^2} < c\), so on implying this to the above expression we get:
\( \Rightarrow {b^2} < c < 3c\)
$\Rightarrow {{b}^{2}}-3c<0$
On multiplying both side with $4$, so that we check the sign of the discriminant we get:
$\Rightarrow 4\left( {{b}^{2}}-3c \right)<0$
Since the discriminant we got is less than zero so ${f}'(x)$ will be greater than zero.${f}'(x)>0$.
$\therefore $ The $f(x)={{x}^{3}}+b{{x}^{2}}+cx+d$ is 4) is strictly increasing if the condition is \(0 < {b^2} < c\).

Note:To solve the problem we need to remember the formulas of the differentiation. For the function to be continuous strictly increasing the differentiation of the function should be greater than zero while for decreasing it should be less than zero.