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azureus88

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Find the gradient of the tangent to the curve at y=f<SUP>-1</SUP>(x) at the point (11,-1) where f(x)=x^3 - 12x.

i keep gettin the wrong answer for some reason. its meant to be -1/9. can someone help out?
 

lolokay

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if you have an inverse function, the x and y change places, since y=f-1(x), x = f(y)
dy/dx at x=-1 in f(x) is 3-12 = -9. In the inverse function, this is dx/dy, so dy/dx is 1/(-9) = -1/9
 

azureus88

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Did u do the question where ABC is a triangle with vertices A(2,5) B(-1,2) C(3,-1) and X divides AC in the ratio m:n and Y divides AB in the ratio m:n. Determine XY/BC?

is the answer really just m? i got m/(m+n)
 

lolokay

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yeah it's m/(m+n). they must have taken m+n to be 1 for some reason (if m=2, n=3 for example, obviously XY is not twice as big as BC: it is 2/5 the size)
 

azureus88

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ok thanks for all the replies. last question for today...determine the values of m such that e^kx = mx will have 2 solutions, 1 solution and 0 solutions. i no that its m =ke for one solution but wat bout for the 2 and 0? any explanation would be nice.
 

lolokay

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well there's one intersection at m=ke, so if m > ke there would be 2, if 0 =< m < ke there would be 0. drawing a graph may help in understanding it: y=e^kx lies above y=kex for all values except one, so if the line increases in gradient it will lie above the exponential for a range of values, and if the gradient of the line is decreased it will lie below the exponential for all values

also, if m is negative, there would only be one intersection

also this assumes k is positive...
 
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