If $ \frac {dy}{dx} = 2 - \frac {3}{t^2 + 1} $, prove that $ -1 \lt \frac {dy}{dx} \lt 2 $ .

Question

Given $ x = t - \frac {1}{t} $ and $ y = 2t + \frac {1}{t} $

From the graph, the range could be visualised. However, the point (0, -1) is defined. Is the question wrong?

Beside using graphing method, is there any alternatives to do the reasoning for the range ?

Answer 1

Hint : Minimum value of, $$ 2-\frac{3}{t^2+1}$$

is at $t = 0$, where the values is $$\min = 2-\frac{3}{0^2+1} = -1$$ But since $x$ is not defined for $t=0$ , $\frac{dy}{dx} \gt -1.$

Similarly , as $\lim_{x \rightarrow \infty} , $ we have maximum value

$$\max = 2-0 =2$$

Hence you can show that : $$-1 \lt \frac{dy}{dx} \lt2$$

Answer 2

$$\frac{(2-y')}{3}=(1+t^2)^{-1}.$$ Let us use $$0<(1+t^2)^{-1} < 1 \implies 0<\frac{2-y'}{3} \le 1\implies -1< y' < 2.$$