1. Derivative Qn

Please do part(ii) of the qn.Thanks

2. Originally Posted by maybeline9216
Please do part(ii) of the qn.Thanks
did you find where the first and second derivative are positive?

do that and you can find where both are positive at the same time

3. Originally Posted by Jhevon
did you find where the first and second derivative are positive?

do that and you can find where both are positive at the same time
How to find?? Just let dy/dx >0; (d^2y/dx^2)>0?? then find the range of values?? But the x are all cancelled out, leaving a constant number only...

4. Originally Posted by maybeline9216
How to find?? Just let dy/dx >0; (d^2y/dx^2)>0?? then find the range of values?? But the x are all cancelled out, leaving a constant number only...
i will do it for the first derivative. i leave it to you to do the second.

we wish to find where $\displaystyle \frac {x^2 - 4x}{(x - 2)^2} > 0$

set the numerator and denominator to zero (note that the function is not defined for the denominator = zero)

we have $\displaystyle x^2 - 4x = 0$

$\displaystyle \Rightarrow x(x - 4) = 0$

$\displaystyle \Rightarrow \boxed{x = 0}$ or $\displaystyle \boxed{x = 4}$

also, $\displaystyle (x - 2)^2 = 0 \implies \boxed{x = 2}$

now, put all these points on a number line.

------------------$\displaystyle \circ$--------------$\displaystyle \circ$--------------$\displaystyle \circ$---------------------
.......................$\displaystyle ~~~~~~~0~~~~~~~~~~~~2~~~~~~~~~~~~~4$

note that we cannot be equal to any of these values (hence the unshaded circles above them), since, if x = 2 we are undefined, while if x = 0 or 4, we are equal to zero. we do not want to be equal to zero, we want to be strictly greater than zero.

now, test the regions. that is, pick a point to the left and right of all the values here. so we can try -1, 1, 3, and 5

take turns plugging these into the fraction to see if we have a negative or positive. (note, we only need to worry about the numerator, since the denominator is a square, thus it is always greater than or equal to zero. this means, it will be positive for all values we plug in besides 2. so the sign of our function depends on the numerator. if the numerator is negative, the function is negative (since a negative number divided by a positive number is negative), if the numerator is positive, then our function is positive (since a positive number divided by a positive number is positive).

so plug in the numbers -1, 1, 3, 5 into x(x - 4)

it is positive for -1
it is negative for 1
it is negative for 3
it is positive for 5

these numbers represent all numbers in their intervals. so it means wherever we got a positive result, the function is positive on that interval.

thus, our function is positive for all $\displaystyle x \in (- \infty, 0) \cup (4, \infty)$

now, you do the other

5. "these numbers represent all numbers in their intervals. so it means wherever we got a positive result, the function is positive on that interval.

thus, our function is positive for all "

I dun understand this

6. Originally Posted by maybeline9216
"these numbers represent all numbers in their intervals. so it means wherever we got a positive result, the function is positive on that interval.

thus, our function is positive for all "

I dun understand this
when we put our numbers on the number line, we divided the number line into 4 intervals: $\displaystyle (- \infty, 0)$, $\displaystyle (0,2)$, $\displaystyle (2,4)$ and $\displaystyle (4, \infty)$. do you see that?

now, i pick numbers in those intervals. the sign of the function will be the same for all numbers in each interval. so i picked $\displaystyle -1 \in (- \infty, 0)$, $\displaystyle 1 \in (0,2)$, $\displaystyle 3 \in (2,4)$ and $\displaystyle 5 \in (4, \infty)$, and tested to see what sign the function would be for these values. the numbers i picked are completely random. i just wanted values that it would be easy to work with. since for -1 and 5 i got positive values, i take the intervals they represent, that is, the intervals that contain them. for those values, our function is positive, since the x-values there return positive results when i plug them in.

7. Oh...thks for ur explanation i understand already,

But what to do for d^2y/dx^2?? which is 8/[(x-2)^3].
Only the denominator matters here...
[(x-2)^3] => x=2

Plot a number line. i choose 1 and 5 (rdm)

The function is negative for 1
The function is positive for 5

So the range here is x>2

Together with the range before(for dy/dx): x< 0 x>4

Therefore, the range of values for both 1st & 2nd derivative is x>4

Is my reasoning right??