help me with Q8 and Q6 only.Q7 no need. thank you very much.
Hello, helloying!
The denominator of $\displaystyle y \:=\:\frac{x-2}{x^2+5}$ is always positive.6) Given: .$\displaystyle y \:=\:\frac{x-2}{x^2+5}$, . find $\displaystyle \frac{dy}{dx}$
and determine the values of $\displaystyle x$ for which both $\displaystyle y$ and $\displaystyle \frac{dy}{dx}$ are positive.
The numerator is positive for $\displaystyle x > 2$
. . Hence, $\displaystyle y$ is positive on $\displaystyle (2,\,\infty)$
The derivative is: .$\displaystyle \frac{dy}{dx} \:=\:\frac{(x^2+5)\!\cdot\!1 - (x-2)\!\cdot\!2x}{(x^2+5(^2} \:=\:\frac{4x-x^2}{(x^2+5)^2} $
The denominator is always positive.
The numerator is a down-opening parabola with intercepts $\displaystyle x = 0,\:4$
The graph is positive between the intercepts.
. . Hence, $\displaystyle \frac{dy}{dx}$ is positive on $\displaystyle (0,4)$
Therefore, $\displaystyle y$ and $\displaystyle \frac{dy}{dx}$ are both positive on: .$\displaystyle (2,\infty) \cap (0,4) \:=\:(2,4)$
Thanks but the question ask me to find a range of values for x. the ans is 2<x<5. u showed me how to get 2<x but i still dont know the other one.the ans u told me is maybe too high level. i havent learn how to solve it yet.could u show me how to get the ans required? thanks a lot.
hi there,
an alternative way to think about it is to simply set the function and it's derivative to 0.
When you set derivative to 0, the resulting value(s) are called critical points. You can evaluate the sign of derivative before and after those critical points by simply plugging a random number before or after (whichever is the case) of the critical point.
I suggest you look into it, if you have not already.