Question

The minimum value of the function $f \left(x\right)=x^{3/2}+x^{-3/2}-4\left(x+\frac{1}{x}\right)$ for all permissible real $x$, is

• A. $-10$
• B. $-6$
• C. $-7$
• D. $-8$

Answer 1

The Correct Option is A

\(f \left(x\right)=x^{3/2}+x^{-3/2}-4\left(x+\frac{1}{x}\right)\) \(f \left(x\right)=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{3}-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) \(-4\left[\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}-2\right]\) Let \(\sqrt{x}+\frac{1}{\sqrt{x}}=t\left(x > 0\right)\) Let \(g\left(t\right) = t^{3} - 3t - 4t^{2} + 8\) \(g \left(t\right) = t^{3} - 4t^{2} - 3t + 8\) \(g'\left(t\right)=3t^{2}-8t-3=\left(t-3\right)\left(3t+1\right)\) \(g'\left(t\right)=0 \Rightarrow t=3\left(t\ne -1/3\right)\) \(g''\left(t\right)=6t-8\) \(g''\left(3\right)=10 > 0 \Rightarrow g\left(3\right)\) is minimum \(g\left(3\right)=27-9-36+8=-10\)

A connection between an element of one non-empty set and an element of another non-empty set is all that constitutes a function. If we broaden the idea and try to simplify it, an equation is a function if it produces exactly one value of Y when evaluated at a specific X for any X in the domain of the equation.

A function is a relationship or association between every element of the non-empty set A and at least one element of the other non-empty set B.

As a result, a relationship between set A (the function's domain) and set B (its co-domain) is created that may be referred to as a function.

It can be mathematically transcribed as:

f = {(a,b)| for all a ∈ A, b ∈ B}

• A relation is described as a function if every element of set A has one and only one image in set B.
• A function is a relation from a non-empty set B to the domain of a function is A and no two distinct ordered pairs in f can have the same first element.
• A function from A → B and (a,b) ∈ f, then f(a) = b, where 'b' is the image of 'a' and 'a' is the pre-image of 'b'. Here the set A is called the domain of the function and set B is to be called the co-domain of it.