Chapter 3/ Test 3

the number e

-e is the natural base of logarithms -used in the "natural" exponential function, it's slope is it's value e= limit as h approaches zero of (1+h)^1/2 the derivative of e^x is itself: f(x)=e^x f'(x)=e^x Examples: 1. f(x)=3e^x f'(x)=3e^x *coefficient stays the same, and the derivative of e^x is itself.* 2. y=x^2e^x y'=x^2(e^x) + 2x(e^x) e^x(x^2+2x) *use product rule* (1st term)(2nd term')+(2nd term)(1st term') 3. y=e^x/x^3 y'=3x^2(e^x) - e^x(x^3) x^2e^x(x-3)/x^6 e^x(x-3)/x^4 *quotient rule* (D')(N)-(N')-(D)/D^2* combine like terms factor out x's and reduce

Elasticity of Demand

E(x)= - x(d'x)/d(X)

find a tangent line to the graph with number e in it

Ex: find the tangent line to the graph f(x)=3e^-7x at the point (0,3) *take first derivative* f(x)=3e^-7x f'(x)= -21x^7 *plug in x value in the point to get slope. here it is 0. e^0=1. f(0)-21e^7(0) -21 is slope *plug into point slope formula* y-y1=m(x-x1) y-3=-21(x-0) y=-21x+3

the decay rate, k, and half-life, t, are related by:

KT=LnT r K=ln2/T T=ln2/K Ex: After how long will half of the 1000grams remain? N(t)=1/2N(of 0) ln2/-.028=25.

Elastic Demand

The demand is elastic if E(X)>1. if demand is elastic, then revenue is decreasing.

Unit Elasticity

The demand is unit elastic if E(x)=1. The demand has unit elasticity at its maximum. Finding price at which total revenue is at a maximum: 1. take the first derivative of r(x) 2. set it equal to zero to get critical values. 3. take the second derivative of r(x). if the second derivative is less than 0, if it is negative, the critical value obtained from the first derivative is where the revenue is at its maximum.

Derivatives of Natural Logs

Theorem #1: the derivative of lnx= 1/x for any positive number x ex: log9x 1/x(ln9) ex2: 3lnx 3'lnx 3/x ex3: x^2nx+5x x^2(1/x) + lnx(2x) +5 (product rule on x^2lnx) x+2x(lnx)+5 Theroem #2: the dervative of a natural log of a function is the derivative of the function divided by the function itself 'lnf(x)=1/f(x) x f'(x)= f'(x)/f(x)

Maximized Revenue

Total revenue is maximized at the value(s) for which e(x)=1

exponential decay

dP/dt=-kP where k>0 : this function shows P to be decreasing as a function of time P(t)=P(of zero)e^-kt : this solution shows it to be decreasing exponentially Ex: decay rate: 2.8%/year (K) dN/dT=-.028N (the rate of change of an amount N) a.) find t=0 N(T)=N(of zero)e^-.028t

Total Revenue. Decreasing.

decreasing at those values for which E(X)>1

the function given by f(x) with 0<a<1 is a positive, decreasing, continuous function. as x gets larger, a ^x approaches 0.

f(x)-a^x where 0<a<1 -positive -decreasing

the function given by f(x)=a^x where a>1 is a positive, increasing, continuous function. as x gets smaller, a^x approaches zero.

f(x)=a^x where a>1 -positive -increasing

exponential function

given by f(x)= a^x where x is any real number, a > 0 and a can't equal one. number a is called a base. x will always be positive.

Total Revenue. Increasing.

increasing at those values for which E(x)<1

logarithmic function

log(base a)^x=y a^y=x where a>0 and a can't equal 1 log(base a)^x is the power y to which we raise a to get x #a is the logarithmic base log(10)100=2 → 10^2=100 log(49)7=1/2 →49^1/2 =7 log(5)1/25=-2 →5^-2 logarithmic equation can be thought of as an exponent, it can be turned into an exponential equation.

Natural Logarithms (LN) & their properties

log(base e)^x is the natural log of x and is abbreviated ln(basex) Most properties of basic logs are the same as natural logs. ln(MN)= lnM+lnN ln(M/N)= lnM- lnN ln(a^k)= k(lna) *lne=1* *ln(e^k)=k* logbM=lnM/lnb lnM=logM/loge ex: w/ lne=1 Solve for e^t=40 for t e^t=40 lne^t=ln40 *take natural log of both sides* t=ln40 *since lne=1* t= 3.7

common logarithmic function

log(base x)= log(base 10)x base 10 is understood as the common log function log1000=3 or 10^3 log.0001=-3 or 10^-3 log10= 1 or 10^1 log.01=-1 or 10^-1

basic properties of logarithms

loga(MN)= logaM +logaN Ex: loga6= loga(2x3)= loga2+loga3=.778 loga(M/N)= logaM- logaN Ex: loga2/3= loga2- loga3= -.176 loga(M^k)=K(the exponent) x logaM Ex: loga81= loga3^4= 4(loga3)= 1.907 loga(A)=1 Ex: logroota=logaA^1/2= 1/2(1)= 1/2 loga1=0 Ex: loga1/3= loga1 - loga3= loga1=0= -loga3= -.477 logbM=logaM/logbM

Limited Growth

p(t)= L(the limit)/ L(1-e^-kt) for k>0 which gives us p(t)=L(1-e^-kt)`

Inelastic Demand

the demand is inelastic if E(X)<1. If the demand is inelastic, the revenue is INCREASING.

a function y=f(x) satisfies

the derivative of ky or f'(x)=k x f(x) Theorem: the derivative of dy/dx is ky or f'(x)=kf(x) if and only if yce^kx or f(x)=ce^kx where c is a constant

The derivative of loga(x)

the derivative of loga(x) is equal to 1 over the natural log of a multiplied by 1 over x 'loga(x)= 1/lna x 1/x Ex: a. y=log8x y' 1/ln8 x 1/x b. logx y'=log10x y'=1/ln10 x 1/x c. log3(x^2+1) y'=1/ln3 x 1/x^2+1 x 'x^2+1 y'= 1/ln3 x 1/x^2+1 x 3x y'= 2x/(ln3)(x^2+1) d. x^3log5x y'= x^3 x 1/ln5 x 1/x + log5x x 3x^2 (product rule) y'= x^2/ln5 + 3x^2log5x

Exponential growth rate

the exponential growth rate K, and the doubling time T, are related by k=ln2/t t=ln2/k ex: P(of 0)=6.0400 k=.016 or 1.6%(move decimal two places) t=ln2/.16=43.3 (years, months, days) uninhibited growth can be modeled by a differential equation dP/dt=KP whose solutions are p(t)=P(of 0)e^kt uninhibited population growth dP/dt=KP or P'(t)=KP(t) where K>0 given by P(t)ce^kt ex: dp/dt=.07p a) p(t)=P(of 0)e^.07t b)p(1)=100(p of 100)e^.07(1) c)doubling time=ln2/.07=9.9(years, months, days)

the derivative of a^x

to find the derivative of a^x, for any base a, we first express a^x as a power of e. to do this, we recall that logb(x) is the power to which b is raised to get x. *to find the derivative of a^x, we differentiate both sides.* d/dx a^x= d/dx^lna^x=a^x(lna) *THEOREM* d/dx a^x= (lna) a^x ~~The derivative of some number a raised to the x power is the natural log of some number a multiplied by a to the x power. ~~~ Ex: a. 2^x ln2(2^x) .7(2^x) b. 1.4^x ln1.4(1.4^x) c.3^2(x) ln3(3)^2x (multiplied by) '2x ln3(3^2x)(2) =2ln3(3)^2x

the derivative of e to some power is the product of e to that power and the derivative of the power

y=e^8x y'=e^8x times 8 y'=8e^8x the exponent on E stays the same even if the coefficient in front come down ae^x= a x 'e^X (the derivative of some number multiplied by e^x is that number times the derivative of e^x) -can distribute e with the exponent, but e doesn't change, exponents are added together.