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integrating 2



I managed to find the integral of f'(x) in part a), but I'm stuck with part b).

I understand that the gradient of normal to P will be -1/2 and therefore, the gradient of the tangent will be +2.

However, where do I go from there? In the mark scheme, they made f'(x) equal to 2 and then => to 0, but I don't understand why.

Could anyone help out please?
Original post by frostyy


However, where do I go from there? In the mark scheme, they made f'(x) equal to 2 and then => to 0, but I don't understand why.

Could anyone help out please?


Can I please see the mark scheme? I am not sure what you mean by "then ==> to 0".
Original post by frostyy


I managed to find the integral of f'(x) in part a), but I'm stuck with part b).

I understand that the gradient of normal to P will be -1/2 and therefore, the gradient of the tangent will be +2.

However, where do I go from there? In the mark scheme, they made f'(x) equal to 2 and then => to 0, but I don't understand why.

Could anyone help out please?

The normal to the curve through P must pass through a point where the curve has gradient of 2. The expression f′(x)f'(x) gives the gradient of the curve at the point with x-coordinate xx, so solving f′(x)=2f'(x)=2 will yield the x-coordinate of all points on the curve with gradient 2 (if you're careful about how you solve it to avoid spurious solutions).
Reply 3
Avatar for B_9710
B_9710
18
Just solve f'(x)=2
(edited 8 years ago)
Original post by B_9710
To go from there just solve the equations simultaneously.
2y+x=0 and y=f(x)

Not quite. We're merely told that the line 2y+x=02y+x=0 is merely parallel to the normal at P, not that it's necessarily the normal at P.
Reply 5
Avatar for frostyy
frostyy
OP
17
I solved the question myself. I had to make f'(x) = 2 (since that's what the gradient for the tangent was) and then just solve for x to find its value, which was 3/2.

Thanks anyway
Reply 6
Avatar for B_9710
B_9710
18
Original post by Farhan.Hanif93
Not quite. We're merely told that the line 2y+x=02y+x=0 is merely parallel to the normal at P, not that it's necessarily the normal at P.


oh yeah. Didn't read it properly.

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