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.

    \(\displaystyle \lim_{x\to 0} f(x)=1,\ \lim_{x\to 1^-} f(x)=-\infty,\ \lim_{x\to -1^+} f(x)=-\infty\).

Difficulty level: Easy task (using definitions and simple reasoning)
Routine calculation training
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