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Calc Mid 1

# Calc Mid 1 - 25(0.8)^3 3.01-3 12.7715-12.8/3.01-3 =-2.85...

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1.7 EXPONENTIAL GROWTH AND DECAY DOUBLING time p= Po a^t a^d = (Po a^t) (2) = 2P Population Po e^kt 21.2= 19.5e^k2 21.2/19.5 = e^k2 take ln both side = ln(21.2/19.5)= 2k = k= 1/2ln(21.2/19.5) Half life Q= Qo e^-.025 t Q = Qo/2 so Qo/2= Qo e^-.025t Ln(1/2) = -.025t then t= ln(1/2)// -.025 1.8 FUNCTION AND CHANGE Y=x^2 and y= (x-2)^2 shifts on x axis to right Y=x^2 and y=x^2 +4 shifts on y axis up Y=f(x) and y= 3f(x) streches on y axis up and down 3X times and y= -2f(x) flips it over x Axis and streches up and down 2 times 1.9 PROPORTIOALITY Y= kx kis constant Proportional: P= f(w) = 1.4w price =p \$1.4 per pound Inv Proportional: Speed v=50 (1/t) = 50/t v= speed 50 = miles Power function : Q(X)= k * x^p W=k/r^2 = kr^-2 p = -2 2.1 ISTANTANEOUS VELOCITY t(sec) y= s(t) (feet) (((s(1.01)-s(1) / 1.01-1 ))) example 90 feet t= 1 velocity == ( (90*1.01)- 90) // 0.01 RATE OF CHANGE Quantity of drug in blood at T mins Q=25(0.8)^T (25(0.8)^3.01

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Unformatted text preview: - 25(0.8)^3) // 3.01-3 12.7715-12.8//3.01-3 = -2.85 DERIVATIVE OF F AT A F’ (2) if f(x) = x^3 ((2.001)^3 - 2^3) // 2.001-2 = 8.012-8//0.001 = 12.0 2.2 DERIVATIVE FUNCTION F’(X) = Instantaneuos rate of chang of f at x F’ >0 increasing over interval F’<0 decreasing over interval F’ = 0is constant 2.3 INTERPREATATIONS OF THE DERIVATIVE Estimate value of function X dollars on ads and y subscriptions F(250) = 180 and F’ (250) =2 means dy/dx = 2 when \$250 if it increase by \$1 subscriptions will go up by 2 2.4 SECOND DERIVATIVE F’’> 0 Means F’ increase so is concaveing up F’’ < 0 Means F’ decrease so is concaving down 2.5 MARGINAL COST AND REVENUE Marginal cost = MC = C’(q) Marginal revenue = MR = R’(q) Marginal cost is smallest when q’s slop is closest to ZERO Q is quantity on the x axis and P is price on y axis MAX profit difference between R and C when R is higher...
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