Question

I am currently working with Applications of Derivatives, and need clarification on something. The problem:
"y = x^2 - x - 1; Determine the following:
1. Inflection Point(s)
2. All Minima & Maxima
3. Intervals where graph is Increasing &
Decreasing
4. Intervals of Concave Ups & Downs"
So, I took the 2nd derivative of the equation, to determine the Inflection Point and got a single integer- no term containing "x". So what's the I.P.? :)

Answer 1

it is a quadratic, you don't need calculus for any of this

Answer 2

Sorry for the abbreviations and poor grammar towards the end there- O.S. wouldn't let me type any more characters.

Answer 3

\[y = x^2 - x - 1\] is a parabolas that opens up

Answer 4

the minimum value is the second coordinate of the vertex
since it is a parabola it is always concave up

Answer 5

the second derivative, if you really want it, is \(y''=2\) a positive constant
that tells you it is always concave up, but you knew that already

Answer 6

oh also that tells you there is no inflection point, it never changes concavity

Answer 7

So... Like literally the inflection point doesn't exist?

Answer 8

Oh, I get it!!!

Answer 9

not to belabor the point, but you know all about quadratics without any calculus at all
it is decreasing unit you get to the first coordinate of the vertex, which in your case is \(\frac{1}{2}\) then increasing from there on

Answer 10

Yep, I get the basic structure of parabolas... I guess I just didn't know what to do, on paper, with just 2 since that derivative is normally, well, supposed to be set equal to zero. :) So in this case... is 2 the y-value of the coordinate for the Absolute Minimum?

Answer 11

no

Answer 12

first coordinate of the vertex is \(-\frac{b}{2a}\) which in your case is \(-\frac{-1}{2\times 1}=\frac{1}{2}\)

Answer 13

second coordinate of the vertex is what you get for \(y\) when you replace \(x\) by \(\frac{1}{2}\)

Answer 14

Ahhh... I wish I had you for a teacher.

Answer 15

Thank you very much!