Consider the curve y=ln(3x-1).let p be the point on the curve where x=2.

1 answer:

4 0

A. Yes you are correct that the gradient at any point is 3/(3x-1). However at point P it would be 3/(3*2-1)=2/5
b. The gradient of the normal would therefore be -5/2
We can use the general formula of an equation to get y-ln(5)=-5/2 (x-2)
Now multiply both sides by 2 to get:
2y-2ln(5)=-5x+10
Now when it crosses the x axis we know that y=0 therefore:
5x=10+2ln(5)
Therefore:
x=2+2/5 ln(5) when y=0
You could find an estimate of this number to be 2.64 (3sf) but this might not be sufficient

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