If y=-3 then it looks like a stationary point, i.e where y'=0, that will help.
Ok, I'm doing a packet for math on derivatives and extrema and this is a problem where I honestly have no idea to start. I know what a curve, derivative, extrema, and tangent line together, but in this problem I don't know how to link them all together.
(noncalculator) the curve with derivative has y = -3 as a tangent line.
at what point is the line tangent to the curve? determine if the point that you found is a relative maximum point, relative minimum point or neither for the curve. Justify your answer.
Any help is appreciated. Thanks!!
As mentioned, the slope of the line y = -3 is zero, thus the slope of the function described by the differential equation will be zero when y = -3. So set your derivative equal to zero and let y = -3 and solve for x. (As you will see, you really don't need to know what y is...)
Then, to determine the nature of the function's behavior at that point, use the second derivative test, using the quotient and chain rules, and substitute into your formula for y'' the values for x, y, and y' to find the sign of y'' at the tangent point.
Ok, plugging -3 in for y I get I then multiply both sides by -1 and do some further algebra to get x=-3.
Assuming this is correct, I then plug in x = -3 to the second derivative, which I began to evaluate but then ran into some complications
1. there is a y in the first derivative, does this mean I must do implicit differentiation?
2. Do I have to use chain rule for (3-x) and (y+2) even though they're not raised to any power?
Thanks