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# integrating 2 Watch

1. 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?
2. (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".
3. (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 gives the gradient of the curve at the point with x-coordinate , so solving 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).
4. Just solve f'(x)=2
5. (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 is merely parallel to the normal at P, not that it's necessarily the normal at P.
6. 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
7. (Original post by Farhan.Hanif93)
Not quite. We're merely told that the line 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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