Question

How do you determine whether the function ${{h}^{'}}\left( x \right)=\dfrac{{{x}^{2}}-2}{x}$ is concave up or concave down and its intervals?

Answer 1

The above mentioned problem is a simple example of differential calculus and graph theory. In problems like these where we have ${{h}^{'}}\left( x \right)=\dfrac{{{x}^{2}}-2}{x}$ , the first thing that we need to do is take $f\left( x \right)=\dfrac{{{x}^{2}}-2}{x}$ . Now on differentiating this equation, and equating this to zero, we find the points of maxima and minima. We further differentiate the equation and put the values obtained earlier to check for the points of maximum or minimum. Equating the second derivative to zero, we find the points of inflections, points where the graph changes its nature.

Complete step by step solution:
Now, starting off with the solution to the given problem, we firstly write,
$f\left( x \right)=\dfrac{{{x}^{2}}-2}{x}$ On further simplification we can write,
$f\left( x \right)=x-\dfrac{2}{x}$
Now differentiating $f\left( x \right)$ with respect to x we get,

& \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( x-\dfrac{2}{x} \right)}{dx} \\\ & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d\left( x \right)}{dx}-\dfrac{d\left( \dfrac{2}{x} \right)}{dx} \\\ & \Rightarrow {{f}^{'}}\left( x \right)=1+\dfrac{2}{{{x}^{2}}} \\\ \end{aligned}$$ Now equating $${{f}^{'}}\left( x \right)$$ to zero, we get, $${{f}^{'}}\left( x \right)=1+\dfrac{2}{{{x}^{2}}}=0$$ Now, from the above equation we try to find out the value of $$x$$ to find the points of maxima and minima. Thus we write, $$\dfrac{2}{{{x}^{2}}}=-1$$ Now, cross multiplying both sides we get, $$-{{x}^{2}}=2$$ On rearranging we get, $${{x}^{2}}=-2$$ From this equation we can clearly understand that there are no real values of $$x$$ , and which means there are no real points of maxima or minima. This also denotes that the function $$f\left( x \right)$$ is either strictly increasing or strictly decreasing. We have the theory that if, $${{f}^{'}}\left( x \right)>0$$ then the function $$f\left( x \right)$$ is a strictly increasing function and if $${{f}^{'}}\left( x \right)<0$$ then the function $$f\left( x \right)$$ is a strictly decreasing function. Keeping in mind of this equation $${{f}^{'}}\left( x \right)=1+\dfrac{2}{{{x}^{2}}}$$ , we can thus clearly say that $${{f}^{'}}\left( x \right)>0$$ for any value of $$x$$ . Thus we can confidently say that $$f\left( x \right)$$ is a strictly increasing function. Now $${{f}^{''}}\left( x \right)=-\dfrac{4}{{{x}^{3}}}$$ , we thus clearly understand that for $$x>0$$ , $${{f}^{''}}\left( x \right)=-\dfrac{4}{{{x}^{3}}}<0$$ which means that the graph of this function will be concave down for $$x>0$$ and that for $$x<0$$ , $${{f}^{''}}\left( x \right)=-\dfrac{4}{{{x}^{3}}}>0$$ which means that the graph of this function will be concave up. **Note:** For these types of problems involving graph theory and differentiation, we need to keep in mind all the above concepts as discussed. Functions like this are also easy to plot and thus, it can be declared as concave up or down just by looking at the graph.