Question

y=x+(1/x). can we find any other maxima or minima at any other point except x=1 and -1.i sketch the graph , at x=1 there is a minima and x=-1 there is a maxima . but can not understand from graph why the maxima is less than the minima.

Answer 1

Answer on Question #50806 – Math – Differential Calculus | Equations

Question

$y = x + (1/x)$. Can we find any other maxima or minima at any other point except $x = 1$ and $-1$? I sketch the graph, at $x = 1$ there is a minima and $x = -1$ there is a maxima, but cannot understand from graph why the maxima is less than the minima.

Solution

If $y^{I}(x_{n}) = 0$ and $y^{II}(x_{n}) < 0$, then $y(x)$ has a relative maximum at $x_{n}$.

If $y^{I}(x_{n}) = 0$ and $y^{II}(x_{n}) > 0$, then $y(x)$ has a relative minimum at $x_{n}$.

See http://en.wikipedia.org/wiki/Second_derivative_test.

Function $y = x + \frac{1}{x}$ is not defined at $x = 0$.

$$y = x + \frac{1}{x}$$ $$y^{I} = 1 - \frac{1}{x^{2}} = \frac{x^{2} - 1}{x^{2}}; \quad y^{I} = 0 \Rightarrow \begin{bmatrix} x_{1} = 1 \\ x_{2} = -1 \end{bmatrix}$$ $$y^{II} = \frac{2}{x^{3}}$$

Hence, the critical values are $x_1 = 1$ and $x_2 = -1$.

Now let us compute $y^{II}(x_n)$:

$$y^{II}(x_1) = y^{II}(1) = \frac{2}{1^{3}} = 2 > 0, \quad \text{so } y(x) \text{ has a relative minimum at } x_1 = 1$$ $$y^{II}(x_2) = y^{II}(-1) = \frac{2}{(-1)^{3}} = -2 < 0, \quad \text{so } y(x) \text{ has a relative maximum at } x_1 = -1.$$

Global minimum and maximum are $-\infty$ and $+\infty$ respectively.

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