consider this equation, $$y=x(400-x)$$the second derivative of this equation is $$y''=-2$$ As far as I know, a negative sign in the second derivative indicates the curve will concave down. As it is a constant I think it says that the curve concaves down all the time. Which means the tangent line will always lie above the function's graph.
QUESTION 1: How can second derivative define concavity? Surely it's not like mathematicians came up with this out of nowhere.
QUESTION 2: What is the significance of knowing where the graph concaves up or down? There may be many but I just to know the absolute basic ones.
EDIT: I messed up badly with concavity last time. But wikipedia made my views clear about concavity. So I edited my questions.
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$\begingroup$Negative second order derivative means the first order derivative is decreasing (doing down).
A negative first order derivative means the curve is decreasing (going down)
In our case, the second derivative is negative, which means the first order derivative is decreasing - it is still positive for $x<200$; as long as first order derivative is positive, the curve is going up.
$\endgroup$ $\begingroup$The 2nd derivative is tells you how the slope of the tangent line to the graph is changing. If you're moving from left to right, and the slope of the tangent line is increasing and the so the 2nd derivative is postitive, then the tangent line is rotating counter-clockwise. That makes the graph concave up.
So the 2nd derivative tells you which way the tangent line is rotating; either counter-clockwise (positive) or clockwise (negative.) This rotation tells you concavity.
BTW, and inflexion point is when the tangent line changes rotational direction.
$\endgroup$ $\begingroup$Re Q1: The second derivative being nonpositive is a sufficient condition for concavity. So if this happens, you are lucky because then your function is concave. (The function $y = -|x|$ is also concave, but it is not even differentiable.)
Re Q2: The power of concavity is that if you encounter a critical point, where the derivative is equal to zero, then you know you have found a global maximizer. Very convenient for Economics problems etc. So if you want a global (or local, it is the same) maximizer for a concave function, look for $x$ where $y'(x)=0$.
$\endgroup$ $\begingroup$Let $$y = x \cdot (400-x) = 400 \cdot x - x^2$$
$$y' = 400 - 2\cdot x$$
$$y'' = -2$$
This just shows that the slope of the curve is reducing.as you increase $x$
$\endgroup$ $\begingroup$To add to @B.Goddard's answer, we can think write the first derivative as:
$$ y' = \tan \theta(x)$$
Differentiating this,
$$ y'' = \sec^2 \theta \frac{d \theta}{dx}$$
Notice that since $ \sec^2 \theta \geq 0$, the sign of $y'' $ is controlled by $ \frac{ d \theta}{dx}$.
If $ \frac{d \theta}{dx} \geq 0 $ , it means that as you move to the right, the tangent line is rotating counter clockwise.
If $ \frac{ d \theta}{dx} \leq 0 $ , it means that as you move to the right, the tangent line is rotating clockwise.
We can think of the tangent line to a curve, doing a translational + rotational motion as we move along the curve to further understand this.
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