C3 numerical methods question

Doing this question instead of making both y terms equal I subbed the curve equation into the normal equation. & I used this for f(x) but when I put the value of of x =1 I get -0.195... Which isn't on the markscheme. I don't get why it doesn't work. Although I still get a sign change I don't have the same values why. Take a look


Question


My working


I got f1 as -0.195...


Part b thanks in advance


Posted from TSR Mobile

Can you show the substitution you made? Surely you have the normal equation in the form y=mx+c, so wouldn't it just be easier to equate the two equations?

I wasn't told to leave it in the form y = mx + c so I left it as 4y = x - 14
As y = 2x - 3ln(2x + 5)

4(2x - 3xLn(2x + 5)) = x - 14
8x - 12xLn(2x + 5) = x - 14
(7x + 14) - 12xLn(2x + 5) = 0

So make this f(x) sub in x=1 and I didnt get the same answer which was 0.54..... or something

Sub 1 into that f(x) you get -2.350....

What I actually made f(x) was

(7x + 14) / 12 - Ln(2x + 5) = 0

But if you see the photo I just left it as (7x + 14) / 12 = Ln (2x + 5) but I took it away when making mty calculations and got f(1) as -0.195.... although I got the sign change. My values were not the same as the mark scheme

I don't know if you just typed it up incorrectly, but your substitution is wrong (see bolded bit in quote); it should be:

As y = 2x - 3ln(2x + 5)4(2x - 3Ln(2x + 5)) = x - 14

Yeah it was just a typo as I got rid of it later on, I cant see what about what I'm doing is wrong and hence why I didnt get the same answer as the markscheme.

You're not doing anything wrong then, you just used a different substitution. You still got a change of sign for f(1) and f(2) so you know that the root q is 1 < q < 2. I'm not too sure what you are asking.

Thanks what I thought was irrespective of the subsitution I did to get F(x) it should still be worth the same value when I substitute values hence my f(1) should be the same as the f(1) in the book regardless of how I or they substituted