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Finding x value knowing the gradient from f'(x)



I managed to do part a) without much issue, but part b) I did with by substitution, which was not the method on the mark scheme and apparently, not the right answer. Where did I make the mistake in my workings?

I think you put x=3 into f(x) so f(3)=....

Don't no exactly until I do the whole question. If that not right let me no I will do the question for you
Reply 2
Avatar for Zacken
Zacken
22
Original post by frostyy
...


You were doing perfectly fine right up till t+9t=10\displaystyle t + \frac{9}{t} = 10.

To get this into a quadratic, you need to multiply both sides by tt which gets you t2+9=10t⇒t2−10t+9=0t^2 + 9 = 10t \Rightarrow t^2 - 10t + 9 = 0 which gives you your values of tt. I'm not sure what you seem to have done.
Reply 3
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Zacken
22
Original post by Aaliyah.97
I think you put x=3 into f(x) so f(3)=....

Don't no exactly until I do the whole question. If that not right let me no I will do the question for you


It's not right. You don't know what f(x)f(x) is. Also don't bother doing the question for him, full solutions are against forum guidelines.
Original post by Zacken
It's not right. You don't know what f(x)f(x) is. Also don't bother doing the question for him, full solutions are against forum guidelines.


Ooh okay I didn't know that I'm knew here so sorry
Don't you just do the original equation they gave you (as it is the differentiated equation) equal to 10 and the two values of x you find is your answer
Reply 6
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Zacken
22
Original post by Aaliyah.97
Ooh okay I didn't know that I'm knew here so sorry


No problem! :-)
Reply 7
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Zacken
22
Original post by Student232
Don't you just do the original equation they gave you (as it is the differentiated equation) equal to 10 and the two values of x you find is your answer


Yes. That's precisely what he's done, except for a small slip when trying to solve his quadratic in x\sqrt{x}.
Reply 8
Avatar for frostyy
frostyy
OP
17
Original post by Zacken
You were doing perfectly fine right up till t+9t=10\displaystyle t + \frac{9}{t} = 10.

To get this into a quadratic, you need to multiply both sides by tt which gets you t2+9=10t⇒t2−10t+9=0t^2 + 9 = 10t \Rightarrow t^2 - 10t + 9 = 0 which gives you your values of tt. I'm not sure what you seem to have done.


Thanks so much! I tried to get rid of the (9/t) by multiplying both sides by 9. Not sure why I thought of that, but it didn't occur to me that that was were the issue was lying. Anyway, thanks a lot :smile:
Reply 9
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Zacken
22
Original post by frostyy
Thanks so much! I tried to get rid of the (9/t) by multiplying both sides by 9. Not sure why I thought of that, but it didn't occur to me that that was were the issue was lying. Anyway, thanks a lot :smile:


No problem! :smile:

You might have been able to avoid this issue by having done f′(x)=10f'(x) = 10 straight away; that is:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{x+ 9}{\sqrt{x}} = 10 \iff x+9 = 10\sqrt{x} \iff x - 10\sqrt{x} + 9 = 0\end{equation*}



Then the substitution t=xt = \sqrt{x} gives you

Unparseable latex formula:

\displaystyle[br]\begin{equation*}t^2 - 10t +9 = 0\end{equation*}



But your method was essentially the same and just as nice, really!
(edited 8 years ago)
Original post by Zacken
Yes. That's precisely what he's done, except for a small slip when trying to solve his quadratic in x\sqrt{x}.


Sorry I didn't look at the working haha only the question :s-smilie:
Reply 11
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Zacken
22
Original post by Student232
Sorry I didn't look at the working haha only the question :s-smilie:


Yeah, don't worry, I'm guilty of doing that far too often. :lol:

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