ext 2 question (1 Viewer)

[kony]

by considering the points of intersection of the curve y = x³ - x + 2 and the line y = mx, show that there is only one tangent to the curve which passes through the origin

any ideas?

 

[modezero]

intersection eq:
x^3-(m+1)x+2=0

all tangents through the origin will intersect the cubic twice, so the equation above has only 2 roots when y=mx is a tangent. In particular one of these roots is a double root (point of tangency). Hence when you differentiate, the equation will still have the same root.

ie 3x^2-(m+1)=0 has one root the same as the first eq

Solve the two equations simultaneously to find m and it only has one possible value (m=2)

 

[ssglain]

Basically, the two curves are required to have the same gradient at their point of intersection.

If you were to do this algebraically:

Let f(x) = x³ - x + 2 and g(x) = mx

For POI: x³ - (m + 1)x + 2 = 0 ...(1)

Also, f'(x) = g'(x) -> 3x² - 1 = m ...(2)

Solve simulateously. The POI turns out to be (1, 2) and m = 2.

I'm not really sure how to verbally express an argument solely based on graphical properties. Anyone else?

 

[kony]

turned out to be quite easily solved. for some reason, when i did the very long method of differentiating f(x) and then solving f'(x) = 0 and then subbing the two roots back into f(x) = 0, i got 2 values of m. [now i think the other m is outside the domain of m]

but yeah, this way only has 1 m. thanks soph. and thanks modezero.