Solving 2nd Derivative Rule: f"(x)=-3(x^2+3)/(x^2-9)^3

[Megz27]

so to find when a graph is concave up or down you can use the 2nd derivative rule, but what happends if we have f"(x) = -3(x^2 +3)/(X^2-9)^3

what I am thinking is that we are looking for f"(x)=0 so x^2+3= O but that would give us x=(-3)^1/2...

any thoughts?

 

[slider142]

Your numerator is always negative, so f''(x) is never 0 (for real x). It is undefined at the points -3 and 3. If we look at the denominator, we see that on the open interval (-oo, -3), the change in the first derivative is negative and decreases without bound, while on the open interval (-3, 3) the change in the first derivative is positive and increases without bound. As the first derivative corresponds to the slopes of tangent lines to the graph, you can see why on the interval (-oo, -3) the graph is concave down while on the second interval, the graph is concave up (Start with an arbitrary tangent line and rotate it along a curve in the direction the second derivative gives).
Try to find the behavior of the graph of f as x decreases or increases without bound and on the third interval (3, oo).

 

[Megz27]

how can we prove that tho? Cause wouldn't you have to find what x is when f '' (x) = O?

 

[slider142]

how can we prove that tho? Cause wouldn't you have to find what x is when f '' (x) = O?

f''(x) is the derivative of the function f'(x), and thus inherits all the theorems about studying the behavior of f using f' (In this case we are studying the behavior of f' using f'').
As to finding when f''(x) = 0, that is only one way in which we can find where the sign of f'' changes. Another way the sign of f'' can change is if f''(x) is undefined when the sign changes, as f'' may not have removable discontinuities at these points. The sign of f''(x) is all we are concerned about when it comes to concave up or down. Note that f''(x) = 0 does not necessarily mean that the sign of f'' has changed; it may just be a minimum or maximum point for f'' so you still have to check the sign of f'' on both sides of that point. Ie., If f(x) = x⁶, then f''(0) = 0, but the graph never changes concavity. It is just "very flat" in the neighborhood of 0.

 

[Megz27]

thanks