For a positive real number $x$, which point has most steepest slope for 1/x??

Question

For a positive real number $x$, which point has most steepest slope for the function $y=\frac{1}{x}$? How can we obtain that?

Answer 1

$$\frac{d}{dx}\bigg(\frac{1}{x}\bigg)=-\frac{1}{x^2},x\not=0.$$So,$$\lim_{x\to 0+}-\frac{1}{x^2}=-\infty,$$$$\lim_{x\to \infty}-\frac{1}{x^2}=0.$$

Answer 2

Let $y = \frac 1x$

Then the slope of the line $y = \frac 1x$ is given by $\frac{y}{x}.$

$$\text {slope} = \frac {\frac 1x}{x}$$ $$\text{slope} = \frac {1}{x^2}$$

To maximize the slope , we have to maximize the function $\frac 1{x^2}$ , which increases as $x \rightarrow 0$.

Hence $\max \text{slope}$ is as $x\rightarrow0$.