Consider the graph at the right of y = f(x). Where is dy/dx pos., neg., 0, or un

Question

Consider the graph at the right of y = f(x). Where is dy/dx pos., neg., 0, or undef.? What about d^2 y/dx^2? On what intervals is the func. INC, DEC, CCU, and CCD? Where does the function have any rel. max. and min.? POI's? Now suppose the graph above is the derivative of some other function g, i.e. g'(x) = f(x). If they can be determined, on what intervals is g INC, DEC, CCU, and CCD? If they can be determined, where does g have any rel. max and min? POI's? Sketch g on the axes above. a) If they exist, find the absolute max. and min. of f(x) = x^3 - 3x +1 on [-3, 3]. What are the CV's? b) Where are there rel. max. and min., if any? Where is there a POI, if any?

Explanation / Answer

Notice the function y = f(x)
is increasing everywhere

So, dy/dx is positive everywhere
negative nowhere
0 nowhere

d^2y/dx^2 :
Clearly the function is concave up from -inf to 0 and
concave diwn from 0 to inf

So, d^2y/dx^2 = pos when (-inf , 0)
= negative when (0 , inf)
= 0 when x = 0

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2)
Inc : (-inf , inf)
Dec : nowhere
CCU : (-inf , 0)
CCD : (0 , inf)

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3)
No rel max or min

POI : (0 , 0)

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4)
g(x) = integral of f(x), as in the area of f...

g increasing :
g' = f

So, g' > 0 when f > 0
So, g increasing : (0 , inf)

g d ecreasing : (-inf , 0)

g'' = f'
Now, g'' > 0 when f' > 0, which is everywhere

So, g is CCU everywhere
And g is CCD nowhere

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5)
At x = 0, g changes from dec to inc

So, relative min exists at x = 0
relative max = DNE
POI :
g'' = 0
which means f' = 0 which is nowhere

So, no POI for g

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6a)
f' = 3x^2 - 3 = 0
x = -1 and 1 which both belong to [-3,3]

Now, we have to find f(-1) , f (-3) , f(1) and f(3)

f(-3) = -17
f(-1) = 3
f(1) = -1
f(3) = 19

So, abs max = 19 when x = 3
abs min = -17 when x = -3

CV's are x = -1 and 1

--------------------------------------------------------------------

6b)
rel max = (-1,3)
rel min = (1,-1)

POI :
Double derivaitve ...
f'' = 6x = 0
x = 0

When x = 0, we get y = 1

So, POI at (0,1)

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