209-1-ans - 4[10pts If the interest is compounded...

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MATH 209 First class test October 04, 2006 1. Find the limits [20pts] (a) \ $\lim\limits_{x\to-2} (2x^2+5x-2)$. Solution: 2(-2)^2 +5(-2)-2=-4 (b) \ $\lim\limits_{x\to1}\frac{x^2+x-2}{x-1}$. Solution: numerator is (x-1)(x+2); cancel the factor of x-1 with the denominator. Then we have lim_{x->1}(x+2)=3. 2. Find the derivatives $f’(x)$ of the following [40pts] (a) \ $f(x)=\frac14 x^{-4}+\frac{2}{\sqrt{x}}+4 e^{2x}$, Answer: -x^{-5}-1/(x\sqrt{x})+8 e^{2x} or -x^{-5}-x^{-3/2}+8 e^{2x}; (b) \ $f(x)=(6x^2+4)^4$, Answer (general power rule) 4(6x^2+4)^3 12x= 48x(6 x^2 +4)^3; (c) \ $f(x)=\frac{x^2-4}{x^2+4}$. Ratio rule: (2x(x^2+4)-2x(x^2-4))/(x^2+4)^2=16x/(x^2+4)^2 (d) \ $f(x)=\ln(\sqrt{x+2})$. Chain rule: (\ln x)’=1/x; (\sqrt{x+2})’=(1/2)(1/\sqrt{x+2})(x+2)’
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=1/(2\sqrt{x+2}). The answer is 1/(2(x+2)). 3. For $f(x)=x^3+4x-2$ calculate [20pts] (a) \ the slope of the tangent line at $x=2$, f’(2)=3(2)^2+4=16 (b) \ the equation of the tangent line at $x=2$. f(2)=14; y=f’(2)(x-2)+f(2)= 16(x-2)+14=16x-18.
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Unformatted text preview: 4. [10pts] If the interest is compounded continuously at the rate $r=0.1$ (10\% annually), how many whole years it needs for the money to double? A=P e^{0.1t} = 2P. Taking natural logarithm on both sides, we have 0.1t=\ln 2=0.69. So, t=6.9 or, rounding up, 7 whole years. [Here you seemingly need a calculator to evaluate \ln 2. In actual midterm, I’ll give you a list of possibly relevant numbers AMONG which there will be the desired 5. [15pts] Market studies for a new processor model showed that the demand of it, depending on the price $p$, is $1,000,000-10,000p$. (a) \ Find marginal revenue at $x=400,000$ (b) \ Find average revenue per unit at $x=400,000$. Demand-price equation is x=1,000,000-10,000 p. That is p=100-x/10,000. R=xp=100x-x^2/10,000. Marginal revenue R’=100-x/5,000 is 20 at x=400,000 (b) \ Average revenue is \bar R=R/x=p=100-x/10,000=60 at x=400,000...
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209-1-ans - 4[10pts If the interest is compounded...

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