A curve has the equation y = 3x^3 - 5x + 4
(a) Find the equation of the tangent to the curve at A(1,2)
(b) Find the coordinates of the other point on the curve
at which the tangent is parallel to the tangent at A.
For your information, answer for (a) is y = 4x -2
I am not sure how to solve the second one. But I shall take a
stab at it anyway. Please point out any irrelevant or wrong
parts.
Solution: Since the tangent is parallel to tangent at A,
gradient = 4.
When you differentiate 3x^3 - 5x + 4, you'd get 9x^2
-5
since you know gradient is 4,
9x^2 - 5 = 4
x^2 = 1 OR -1
Since x =1 already belongs to the coordinates of A, the x which
is required in the coordinate we need to find is -1 instead
sub -1 into 3x^3 - 5x +4
y = 6
therefore the other coordinate is (-1,6)
It would be great if someone has an alternative way to solve
this question. But do tell me if this method is valid. Lots of
thanks.
The method you used is valid. You sec 4 this year ah? So fast
learning application of differientation liao ah?
zzz... half of my students only just finish product and quotient
rules... only 1 completed differentiation and geometric
proof...zzz... I think I must hasten the pace...