MATH 1081 Calculus for Social Science and Business – Flashcards

Set equation equal to 0: x²-x-12 = 0 Factor: (x-4)(x+3) = 0 x = 4 and x = -3 these 2 pts represent changes in the graph - so find whether the factor will be positive or negative in each area. Calculate for each factor: (x-4)...........-.|..-..|..+ (x+3)..........-.|..+..|..+ (x-4)(x+3) .+.|..-..|..+ line.....(??..-3....4....+?) This table represents how the factors change based on numbers in that area. for x-4 between negative infinity and -3 i chose -5 - so -5-4 equals -9 and all #s in that area will result in x-4 being negative and for (x+3) use -5, and all numbers between neg infinity and -3 will result in negative result for (x+3) then for multiplying (x-4)(x+3) => - * - = positive, betwix -3 to 4, -*+=-, etc.

2 lines are parallel if and only if they have the same slope or if they are both vertical

2 lines are perpendicular if and only if the product of their slopes is -1, or if one is vertical and the other horizontal

a relationship f defined by y = f(x) = mx + b, for real #s m and b, is a linear function

C(x) = mx + b, the m is the marginal cost and b the fixed cost.

(y?-y?)/(x?-x?)=m where x?,y? and x?,y? are points on the line and m is the slope of the line

y=mx+b m is the slope of the line b is the y-intercept of the line

y-y?=m(x-x?) m is the slope (x?,y?) is a point on the line

vertical line, undefined slope, i.e. 1/0

horizontal line, slope=0

Y=mx+b, slope is m, b is y-intercept: b=y-mx that is to say, given a point on the line, just put in the x, y into the equation and solve to get b - the y-intercept given a point xsubn,ysubn the equation: b=(?y - m(?x)/n

m=(n(?xy)-(?x)(?y))/(n(?x²)-(?x)²)

Revenue = Cost R(x)=C(x)

R(x)-C(x)

F = 9/5C +32

C=5/9(F - 32)

How well the original data fits a straight line: r = (n(?xy)-(?x)(?y))/(?(n(?x²)-(?x)²) ×?(n(?y²)-(?y)²))

a function is a rule that assigns to each element from one set exactly one element from another set

the set of all possible values of the independent variable in a function is called the domain of the function (x), and the resulting set of possible values of the dependent variable is called the range (y)

a graph represents a function if and only if every vertical line intersects the graph in no more than one point

the graph of the quadfunc f(x) = ax²+bx+c is a parabola with its vertex at (-b/2a,f(-b/2a)). the graph opens upward if a>0 and downward if a<0

3 or more points that are in a straight line

* the correlation coefficient measures the strength of the linear relationship between 2 variables * r is between 1 and -1 or equal to 1 or -1 * if r = 1, the least squares line has a positive slope * r = -1 gives a negative slope * if r = 0 there is no linear correlation between the data points (but some nonlinear function might provide excellent fit for the data) * also a r=0 may also indicate that the data fit a horizontal line. * the exact value of the linear relationship depends upon n - the number of data points

No. a positive correlation means that as one of the quantities increases, the other quantity also increases. It does not mean that one of the quantities increasing causes the other to increase. To prove that one of the quantities increasing causes the other to increase would require further research.

a?=1 a??=1/a?

m=0

m is undefined , i.e., 1/0

a

|a| if n is even a if n is odd

??(ab)

??(a/b) b?0

(?*?)?a

??a example 9¹/² = ²?9 = 3

(a¹/?)^m example 27^2/3 = (27^1/3)²=3² = 9

If r is the annual stated rate of interest and m is the number of compounding periods per year, the effective rate of interest is: r subE = ( 1 + r/m)^m -1

y=y?e^(kt) Let y? be the amount or # of some quantity present at time t=0. The quantity is said to grow or decay exponentially if for some constant k, the amt present at time t .

For every positive real number a: a^x = e ^((ln a)x)

a & b > 0, a?1, b?1 log sub a x = (log sub b x) / (log sub b a)

a^y = x

A = Pe^(rt) dollars If a deposit of P dollars is invested at a rate of interest r compounded continuously for t years, the compound amount is A.

As m becomes larger and larger, (1 + 1/m)^m becomes closer and closer to the number e whose approximate value is 2.718281828

n-1

at least n + 1

go up or down

one end goes up and one end goes down

positive

negative

a + b = b + a ab = ba

(a + b) + c = a + (b + c) (ab)c = a(bc)

a(b + c) = ab + ac

-b/2a is the x y = f(-b/2a) (-b/2a,f(-b/2a)) point of vertex

a > 0

a^(m+n)

a^(m-n)

a^(mn)

a^m * b^m

a^m/b^m

log_a x + log_a y

log_a x - log_a y

r * log_a x

1

r

a(x+d)²+e=0 d = b/2a e = c - (b²/4a)

1. First, factor a from 1st and 2nd term a(x²+(b/a)x) + c 2. Then take half of the coefficient of x inside the parans and square it. Subtract a times this number from c a(x²+(b/a)x+[(b/2a)²]) + c -(a*(b²/4a²) 3. simplifying: a(x+b/2a)²+c-(b²/4a)=y Note: so h = -b/2a and k = c-(b²/4a)

1. 3(x²+3x)+1 2. 3(x²+3x+9/4)²+1-(3*9/4) 3. 3(x+3/2)²+1-(27/4) 4. y=3(x+3/2)²-(23/4) vertex of this formula is (-3/2,-23/4)

horizontal asymptote is y=0

no horizontal asymptote

horizontal asymptote is equal to coefficients of highest degree for this example: y=2