Having trouble calculating concave up and concave down?

Question

Here is the question I am having trouble withL

 

Calculate the second derivative of f. Find where f is concave up, concave down, and has inflection points. f(x)= (3x^2) / (x^2 + 49)?
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I figured out the second derivative.
f"(x) = -(294 (3x^2 - 49)) / (x^2 +49)^3
and that the Inflection points are:
x = -4 and 4
However on what intervals are the Concave up and Concave down located? Thank you!

Answer 1

Hi Alison,

 

To determine the concavity of a function, you need to know the sign of the 2nd derivative over the particular intervals between inflection points and discontinuities. This particular function has no discontinuities, so you only have to worry about inflection points as boundaries.

 

If f'' is positive, f is concave up. If f'' is negative, f is concave down.

 

Given inflection points at x ≈ 4 and -4, you get three regions to test:

(-∞. -4), (-4, 4), (4, +∞)

 

In each case, you just need to plug in a value and determine what the sign will be. You don't have to evaluate the function fully.

 

f"(x) = -(294 (3x^2 - 49)) / (x^2 +49)^3

The factor 294 won't change the sign, so you can ignore it. The denominator is always positive, so that has no effect on the sign either. What's left is -(3x^2 - 49)

 

For the first interval, try plugging in -5 (or -6 or -7).

-(3x^2 - 49)  @x=-5

= -(3(-5)^2 - 49)

= -(75-49)

This is a negative number, so on the interval (-∞. -4), f(x) is concave down.

 

For the second interval, x=0 is easy to plug in:

-(3x^2 - 49) @x=0

= -(3(0)^2 - 49)

= +49

On the interval (-4, 4), f(x) is concave up.

 

For the third interval, test x=5 (or 6 or 7), and you'll get the same results at for -5. f(x) is concave down on (4, +∞).

Answer 2

Definition. Let f(x) be a differentiable function on an interval I.

 

(i)
We will say that the graph of f(x) is concave up on I iff f '(x) is increasing on I.

 

(ii)
We will say that the graph of f(x) is concave down on I iff f '(x) is decreasing on I.

 

A point at which a graph changes from being concave up to concave down, or vice versa, is called an inflection point. 

 

Since you have found your inflection points, all you have to do is check each region to see what is happening to the derivative.  You only care about the sign of the second derivative, not the actual value.

 

If the value of the second derivative is positive, then the first derivative is increasing, so that part is concave up.

 

If the value of the second derivative is negative, then the first derivative is decreasing, so that part is concave down.

 

Or, you can look at the graph and say - look, it's cupped downward here, so that's concave down!

 

If I read your f'' right, then substituting a value of zero, the second derivative at X=0 is -(294(-49))/(49)^3).  Minus a minus is plus, so that's a positive number, and that part is concave up. 

 

Substituting +7 and -7, in both cases the (3x^2-49) term will switch signs and no other signs will change, so that's a negative number, and those two parts are concave down.