Derivative Equations

Question

Let f(x)=x^3+3x^2-144x+1 

 

on what interval is f increasing (including the endpoints in the the interval)? Interval of increasing = ----

 

On what interval is f decreasing (including the endpoints in the interval)? Interval of decreasing ---

 

On what interval is f concave downward? ---

 

On what interval is f concave upward? --- 

 

Answer 1

Michelle:

 

f(x) = X³ +3X² - 144X + 1

To find the behaviors you need, you must first derive it. That's simple. Power rule all the way. That gets you:

f'(X) = 3X² + 6X - 144

Set it equal to zero and solve for X:

3X² + 6X - 144 = 0

3(X² + 2X  - 48) = 0

3(X + 8)(X - 6) = 0

X = -8  ,  X = 6

These are the critical values. The function turns horizontal at each of these. To find out what it does before or after each one, we need the second derivative:

f''(X) = 6X + 6

f''(-8) = 6(-8) + 6 = -48 + 6 = -42

f''(6) = 6(6) + 6 = 36 + 6 = 42

This means that at X = -8, it's concave down and at X = 6 it's concave up. How do we know when it changes from one to the other? Inflection point(s). Set the f'' to zero:

6X + 6 = 0

6X = -6

X = -1

Now we know it was concave down from -∞ to -1, and concave up from -1 to +∞. That in turn means it is:

rising from -∞ to -8

falling from -8 to 6

rising from 6 to +

infinity

Answer 2

f(x)=x³+3x²-144x+1

FOC: f'(x)=3x²+6x-144=0; x=-8, x=6 are critical points.

 

f''(x)=6x+6

f''(-8)=-42<0 -->x=-8 is a maximum point

f"(6)=42>0,-->x=6 is a minimum point

 

f(x) increases on the intervals (-∞, -8) and (6, ∞) and decreases

on (-8, 6).

You can define regions of concavity and convexity by using the second derivative:f''(x)=6x+6; it is zero at x=-1. So, you have to see the sign of f''(x) before and after x=-1.