Transcript Derivative by Qiulin He

Let f be defined on an interval, and let x1 and x2 denote points in that interval.
(a) f is increasing on the interval if f(x1)<f(x2) whenever x1<x2
(b) f is decreasing on the interval if f(x1)>f(x2) whenever x1<x2
(c) f is constant on the interval if f(x1)= f(x2) for all points x1 and x2.
Let f be a function that is continuous on a closed interval [a,b] and
differentiable on the open interval (a,b).
(a) If f’(x)>0 for every value of x in (a,b), then f is increasing on [a,b].
(b) If f’(x)<0 for every value of x in (a,b), then f is decreasing on [a,b].
(c) If f’(x)=0 for every value of x in (a,b), then f is constant on [a,b].
Definition. If f is differentiable on
an open interval I, then f is said to
be concave up on I if f’ is
increasing on I, and f is said to be
concave down on I if f’ is
decreasing on I.
If f is continuous on an open
interval containing a value
x0,and if f changes the direction
of its concavity at the
point(x0,f(x0)), then we say that f
has an inflection point at x0,
and we call the point(x0,f(x0)) on
the graph of f an inflection
point of f
Use the graph of the equation y=f(x) in the
accompanying figure to find the signs of
dy/dx and dy/dx and d2y/dx2at the points
A,B , and C.
dy/dx<0 , d2y/dx2>0
dy/dx >0 , d2y/dx2<0
dy/dx<0 , d2y/dx2<0
Use the graph of f’ shown in
the figure to estimate all
values of x at which f has(a)
relative minima , (b) relative
maxima, and (c) inflection
points
2 Because the curve is
turning negative to
positive .
0 Because the curve is turning positive to negative
3 Because the slope of f is changing negative to positive
Find any critical numbers
of the function
g(x) = x2(x2 - 6)
g′ (x) = (x2) ′ (x2 - 6) + (x2)(x2 - 6) ′
g′ (x) = 2x(x2 - 6) + (x2)(2x)
g′ (x) = 4x3 - 12x.
g′ (x) = 4x(x2 - 3).
Since g′ (x) is a
polynomial, it is defined
Remember that a critical
number is a number in the everywhere. The only
numbers we need to find
domain of g where the
are the numbers where
derivative is either 0 or
the derivative is equal to
undefined.
0, so we solve the
equation
4x(x2 - 3) = 0.
The solutions are
These are the only critical
numbers.