Calculus Question (Tangents and Secants and Derivatives)?

Question

F(x) = x^3

The slope of point P passes through point Q. Prove that the slope of the tangent of Q is 4 times the slope of the tangent of P.

How do you do this?

Answer 1

First, let the point P=(t, F(t))=(t, t³). Now, the tangent to F at P will have slope 3t², so in point-slope form it is:

y-t³ = 3t²(x-t)

Now, we want to find the points where this intersects the original function. Obviously P is one such point, so we must find the others. Any point on F will be of the form (x, x³), so substituting this into the equation:

x³-t³ = 3t²(x-t)

Solving for x:

x³-t³ = 3t²x-3t³
x³-3t²x+2t³ = 0

We must factor this. Since we know P is on the line, this means one of the solutions will be x=t. So (x-t) will be a factor. Using synthetic division:

t | 1 0 -3t² 2t³
...... t ....t² -2t³
--------------------
... 1 t -2t² .. 0

So the other factor is x²+tx-2t². Setting this equal to zero and using the quadratic formula to obtain the other solutions:

x=(-t±√(t²+8t²))/2 = (-t±3t)/2
x=t or x=-2t

Thus the only points where the tangent line passes through F are at x=t (which gives us point P), and x=-2t. Thus the x-coordinate of Q must be -2t. The derivative at Q is then 3(-2t)² = 12t² = 4*(3t²), which is four times the derivative at P, which was what we were trying to prove.

Answer 2

you recognize how you won't be in a position to appreciate the fee something is going and it is spectacular region on the comparable time. The by-product is a manner of looking like an aproximate velocity at a set time, by way of fact of this once you % a slope you attempt to discover 2 factors that are very close mutually. There are some very clever aplications for the derivitave yet i won't be in a position to think of of them splendid now.

Answer 3

Re-read, then re-write your question. Start with points don't have a slope....