M119 Review Answers

  1. $$f'(x)=e^{\sqrt x}{1\over\sqrt x}+{3\over x}$$.
  2. $$\lim_{h\to 0}{\ln(2+h)-\ln 2\over h}=f'(2)$$ where $f(x)=\ln x$. Since $f'(x)=1/x$ then the answer is $f'(2)=1/2$.
  3. $$\int_0^{1/5}(25x-x^2)\,dx+\int_{1/5}^{\sqrt 5}(5-x^2)\,dx =-{1\over 2}+{10\over 3}\sqrt 5$$.
  4. There is a relative maximum at $(3,1/6)$ and a relative minimum at $(-3,-1/6)$.
  5. Let $A$ be the area of the poster and $P$ be the area of the printed matter. Then $A=xy=125$ and $$P=(x-4)(y-5)=(x-4)\Big({125\over x}-5\Big)$$. It follows that $P'(x)=-5(x^2-100)/x^2=0$ when $x=10$. This is a maximum because $P'(x)$ is positive for $0<x<10$ and negative for $x>10$. The dimensions of the poster should be $10$ inches wide by $12.5$ inches high.
  6. $\int e^{5x}={1\over 5}e^{5x}+C$. Setting $f(3)={1\over 5}e^{15}+C=5$ yeilds that $C=5-{1\over 5}e^{15}$. It follows that $f(2)={1\over 5}e^{10}+5-{1\over 5}e^{15}$.
  7. Set $u=\ln x$ so that $du=(1/x)dx$. The integral becomes $\int \sqrt u\,du={2\over 3}u^{3/2}+C = {2\over 3}(\ln x)^{3/2}+C$.
  8. $$\lim_{x\to\infty}{3x^2+12x-9\over 5x^2+8}={3\over 5}$$.
  9. Use implicit differentiation. Hence $y'e^x+ye^x+e^y+xe^y y'=0$ so that $$y'=-{ye^x+e^y\over e^x+xe^y}=-{1\over e+1}$$.
  10. Solve for $t$ in $350=100e^{0.08t}$ to get $$t={\ln 3.5\over 0.8}$$.
  11. $\Delta x=1/5$ and $x_1=1$, $x_2=6/5$, $x_3=7/5$, $x_4=8/5$, $x_5=9/5$. Therefore the Riemann sum is $$\Big(1+{5\over 6}+{5\over 7}+{5\over 8}+{5\over 9}\Big) \Big({1\over 5}\Big)={1879\over 2520}$$.
  12. $$y=e^{x^3}$$, $$y'=e^{x^3}(3x^2)$$, $$y''=3xe^{x^3}(2+3x^3)$$. If follows $y$ is increasing on $(-\infty,\infty)$, concave up on $(-\infty,-(2/3)^{1/3})\cup (0,\infty)$, and concave down on $(-(2/3)^{1/3},0)$.
  13. $${1\over 3}x^3+{5\over 2}x^2+17x\Big|_0^5={1135\over 6}$$.
  14. Using properties of logarithms $f(x)=2\ln\big(x^2+5)+5\ln(x^5+7)$. Thus $$f'(x)={2\over x^2+5}(2x)+{5\over x^5+7}(5x^4)$$ and $f'(1)=91/24$.