Shape of a function
Task number: 3053
Investigate the shape of the function \(f(x)=x\sqrt{1-x^2}\).
That is, determine its domain of definition, range, extrema, inflection points, asymptotes; investigate monotonicity, convexity, concavity, behavior at boundary points of the domain of definition. With these characteristics in mind, sketch its graph.
Resolution
The function is odd and continuous.
The domain of definition \(D_f=\langle -1, 1\rangle\).
\(f(x)=0\) for \(x\in\{-1{,}0,1\}\).
\(\displaystyle f'(x)= 1\cdot \sqrt{1-x^2}+ x \cdot \frac1{2\sqrt{1-x^2}}\cdot (-2x)= \sqrt{1-x^2}- \frac{x^2}{\sqrt{1-x^2}} \)
\(f'(x)=0\) for \(x\in\{\frac{-\sqrt2}2,\frac{\sqrt2}2\}\).
On \(\langle-1,\frac{-\sqrt2}2\rangle\cup\langle\frac{\sqrt2}2{,}1\rangle\) the function is decreasing, on \(\langle\frac{-\sqrt2}2,\frac{\sqrt2}2\rangle\) it is increasing.
\(f\left(\frac{-\sqrt2}2\right)=-\frac12\) is a minimum, and \(f\left(\frac{\sqrt2}2\right)=\frac12\) is a maximum. The range \(H_f=\left\langle-\frac12,\frac12\right\rangle\).
\(\displaystyle f''(x)= \frac1{2\sqrt{1-x^2}}\cdot (-2x) -\frac{2x}{\sqrt{1-x^2}} + \frac{x^2\cdot(-2x)}{2\sqrt{1-x^2}^{3}}= -\frac{2x}{\sqrt{1-x^2}} - \frac{x^3}{\sqrt{1-x^2}^{3}} \)
\(f''(x)=0\) for \(x=0\).
On \(\langle-1{,}0\rangle\) the function is convex; on \(\langle0{,}1\rangle\) it is concave. 0 is an inflection point.
\(\displaystyle \lim_{x\to 0} f(x)=1,\ \lim_{x\to 1^-} f(x)=-\infty,\ \lim_{x\to -1^+} f(x)=-\infty\).
