Recent content by drandhawa

Never mind, I see what I did wrong. I should have put the two dy/dx's on the same side, take out dy/dx and then just divide to get dy/dx alone.

what I did was... y^2= 2xy +21 2dy/dx=2(y+x(dy/dx)) 2(dy/dx)= 2y+2x(dy/dx) Divide by dy/dx on both sides dy/dx=2y+2x

i tried to diff it in terms of dy/dx but I don't think i got it right. i got dy/dx to equal 2y+2x

Homework Statement If y^2 - 2xy=21, then dy/dx at the point (2,-3) is ? Homework Equations y^2 - 2xy=21 The Attempt at a Solution I know that I have to differentiate the function. I just do not know how to do it: implicitly or explicitly.

Oh, so because at the min pt is at (3,-9), I just have to shift the graph up so that the min pt would be (3,11). This means that the k value would have to be 20, right?

what di I do about the K?

I plugged in 3 for f(x) and got -9 + k = f(x)

Homework Statement Let f be a function given by f(x)= x^3 - 5x^2 + 3x + K, where k is a constant. Find the value of k for which f has 11 as it's relative minimum. Homework Equations f(x)= x^3 - 5x^2 + 3x + K The Attempt at a Solution First, I found the derivative of f(x), found...

Homework Statement The function y=x(to the forth)+bx(squared)+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b? Homework Equations I think you have to find the derivative, set it equal to 0 and solve for x The Attempt at a...