Math 1710
Topics for first exam
Chapter 1: Limits and Continuity
 § 1:

Rates of change and limits
Calculus = Precalculus + (limits)
Limit of a function f at a point x_{0} = the value the function `should' take at the point
= the value that the points `near' x_{0} tell you f should have at x_{0}
lim_{x®x0}f(x) = L means f(x) is close to L when x is close to (but not
equal
to) x_{0}
Idea: slopes of tangent lines
Figure
lim_{x®x0}f(x) = L does
not care what f(x_{0})
is; it ignores it
lim_{x®x0}f(x) need not exist! (function can't make up it's mind?)
 § 2:

Rules for finding limits
If two functions f(x) and g(x) agree (are equal) for every x near a
(but maybe not
at a), then
lim_{x® a}f(x) = lim_{x® a}g(x)
Ex.: lim_{x®2} [(x^{2}3x+2)/(x^{2}4)] = lim_{x®2} [(x1)/(x+2)]
If f(x)®L and g(x)®M as x®x_{0} (and c is a constant), then
f(x)+g(x)®L+M ; f(x)g(x)®LM ; cf(x)®cL ;
f(x)g(x)®LM ;
and f(x)/g(x)®L/M
provided M ¹ 0
If f(x) is a polynomial, then lim_{x® x0}f(x)= f(x_{0})
Basic principle: to solve lim_{x® x0} , plug in x = x_{0} !
If (and when) you get 0/0 , try something else! (Factor (xx_{0}) out of top and bottom...)
If a function has something like Öx  Öa in it, try multiplying (top and bottom)
with Öx + Öa
Sandwich Theorem: If f(x) £ g(x) £ h(x) , for all x near a (but not
at a), and
lim_{x® a}f(x)lim_{x® a}h(x)L , then lim_{x® a}g(x)L .
 § 4:

Extensions of the limit concept
Motivation: the Heaviside function
Figure
Limit from the right: lim_{x® a+}f(x)L means f(x) is close to L
when x is close to, and
bigger than, a
Limit from the left: lim_{x® a}f(x)M means f(x) is close to M
when x is close to, and
smaller than, a
lim_{x® a}f(x)=L then means lim_{x® a+}f(x)=L and lim_{x® a}f(x)=L
Infinite limits: ¥ represents something bigger than any number we can think of
lim_{x® a}f(x)¥ means f(x) gets really large as x gets close to a
Also have lim_{x® a}f(x)¥ ; lim_{x® a+}f(x)¥ ;
lim_{x® a}f(x)¥ ; etc....
Typically, an infinite limit occurs where the denomenator of f(x) is zero
(although not always)
 § 5:

Continuity
A function f is
continuous (cts)
at
a if lim_{x® a}f(x)f(a)
This means: (1) lim_{x® a}f(x) exists ; (2) f(a) exists ; and
(3) they're equal.
Limit theorems say (sum, difference, product, quotient) of cts functions are cts.
Polynomials are continuous at every point;
rational functions are continuous except
where denom=0.
Points where a function is not continuous are called
discontinuities
Four flavors:
removable: both onesided limits are the same
jump: onesided limts exist, not the same
infinite: one or both onesided limits is ¥ or ¥
oscillating: one or both onesided limits DNE
Intermediate Value Theorem:
If f(x) is cts at every point in an interval [a,b], and M is between f(a) and f(b),
then there is (at least one) c between a and b so that f(c) = M.
Application: finding roots of polynomials
 § 6:

Tangent lines
Slope of tangent line = limit of slopes of secant lines; at (x_{0},f(x_{0}) :
lim_{x® x0}[(f(x)f(x_{0}))/(xx_{0})] Notation: call this limit f^{¢}(x_{0})derivative of f at x_{0}
Different formulation: h = xx_{0}, x = x_{0}+h
f^{¢}(x_{0})lim_{h® 0}[(f(x_{0}+h)f(x_{0}))/h]
Chapter 2: Derivatives
 § 1:

The derivative of a function
derivative = limit of difference quotient (two flavors)
f^{¢}(x_{0}) exists, say f is
differentiable at x_{0}
Fact: f differentiable (diff'ble) at x_{0}, then f cts at x_{0}
h® 0 notation: replace x_{0} with x (= variable), get f^{¢}(x)
new
function
f^{¢}(x)the derivative of f = function whose vaules are the slopes of the tangent
lines to the graph of y=f(x) . Domain = every point where the limit exists
Notation:
f^{¢}(x)[dy/dx][d/dx](f(x)) = [df/dx] = y^{¢} = D_{x} f = Df = (f(x))^{¢}
 § 2:

Differentiation rules
[d/dx](constant) = 0
[d/dx](x) = 1
(f(x)+g(x))^{¢} = (f(x))^{¢}+ (g(x))^{¢}
(f(x)g(x))^{¢}= (f(x))^{¢} (g(x))^{¢}
(cf(x))^{¢}= c(f(x))^{¢}
(f(x)g(x))^{¢}= (f(x))^{¢}g(x)+ f(x)(g(x))^{¢}
([f(x)/g(x)])^{¢}= [(f^{¢}(x)g(x)f(x)g^{¢}(x))/(g^{2}(x))]
(x^{n})^{¢}= nx^{n1} , for n=0,1,1,2,2,3,.......
(( (1/g(x))^{¢}= ggp/(g(x))^{2} ))
f^{¢}(x) is `just' a
function, so we can take its derivative!
(f^{¢}(x))^{¢}= f^{¢¢}(x)(= y^{¢¢} = [(d^{2}y)/(dx^{2})] = [(d^{2}f)/(dx^{2})])
= second derivative of f (=rate of change of rate of change of f !)
Keep going! f^{¢¢¢}(x) = 3rd derivative, f^{(n)}(x) = nth derivative
 § 3:

Rates of change
Physical interpretation:
f(t)= position at time t
f^{¢}(t)= rate of change of position = velocity
f^{¢¢}(t)= rate of change of velocity = acceleration
f^{¢}(t) = speed
Basic principle: for object to change direction (velocity changes sign),
f^{¢}(t)= 0 somewhere (IVT!)
Examples:
Freefall: object falling near earth; s(t) = s_{0}+v_{0} t[g/2] t^{2}
s_{0} = s(0) = initial position; v_{0} = initial velocity; g= acceleration due to gravity
Economics:
C(x) = cost of making x objects; R(x) = revenue from selling x objects;
P = RC = profit
C^{¢}(x) = marginal cost = cost of making `one more' object
R^{¢}(x) = marginal revenue ; profit is maximized when P^{¢}(x) = 0 ;
i.e., R^{¢}(x) = C^{¢}(x)
 § 4:

Derivatives of trigonometric functions
Basic limit: lim_{x®0}[sinx/x] = 1 ; everything else comes from this!
Note: this uses radian measure! lim_{x®0}[sin(bx)/x] = b
Then we get:
(sinx)^{¢}= cosx (cosx)^{¢}= sinx
(tanx)^{¢}= sec^{2} x (cotx)^{¢}= = csc^{2} x
(secx)^{¢}= secx tanx (cscx)^{¢}= = cscxcotx
 § 5:

The Chain Rule
Composition (g°f)(x_{0}) = g(f(x_{0})) ; (note: we
don't know what g(x_{0}) is.)
(g°f)^{¢} ought to have something to do with g^{¢}(x) and f^{¢}(x)
in particular, (g°f)^{¢}(x_{0}) should depend on f^{¢}(x_{0}) and g^{¢}(f(x_{0}))
Chain Rule: (g°f)^{¢}(x_{0}) = g^{¢}(f(x_{0}))f^{¢}(x_{0})
= (d(outside) eval'd at inside fcn)·(d(inside))
Ex: ((x^{3}+x1)^{4})^{¢}= (4(x^{3}+11)^{3})(3x^{2}+1)
Different notiation:
y = g(f(x)) = g(u), where u = f(x), then [dy/dx] = [dy/du][du/dx]
 § 6:

Implicit differentiation
We can differentiate functions; what about equations? (e.g., x^{2}+y^{2} = 1)
graph
looks
like it has tangent lines
Figure
Idea:
Pretend equation defines y as a function of x : x^{2}+(f(x))^{2} = 1
and differentiate!
2x+2f(x) f^{¢}(x) = 0 ; so
f^{¢}(x) = [(x)/f(x)] = [(x)/y]
Different notation:
x^{2}+xy^{2}y^{3} = 6 ; then 2x+(y^{2}+x(2y[dy/dx])3y^{2}[dy/dx] = 0
[dy/dx] = [(2xy^{2})/(2xy3y^{2})]
Application: extend the power rule
[d/dx](x^{r}) = rx^{r1} works for any rational number r
 § 7:

Related Rates
Idea: If two (or more) quantities are related (a change in one value means a change in
others), then their rates of change are related, too.
xyz = 3 ; pretend each is a function of t, and differentiate (implicitly).
General procedure:
Draw a picture, describing the situation; label things with variables.
Which variables, rates of change do you know, or want to know?
Find an equation relating the variables whose rates of change you know or want to
know.
Differentiate!
Plug in the values that you know.
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