Homework 5, due July 20 at beginning of class.

  1. Label the local/global minima/maxima in these graphs, if they exist (the dashed lines indicate the endpoints of the domains of the functions, and a red line indicates an asymptote).

     

  2. Sketch the following functions by determining where they are increasing, decreasing, concave up, concave down, find the vertical, horizontal, and oblique asymptotes (if they exist), and draw them so that they are the correct values at the critical and inflection points (i.e. where $f'(x)$ or $f''(x)$=0).
    1. $5x^{2}+10x$
    2. $x^{3}-3x$
    3. $\frac{1}{1-x^{2}}$
    4. $\frac{x^{2}-1}{x+2}$ (This has an oblique asymptote, and we will talk about this on Monday)

     

  3. Find the maximum and minimum values of the following functions in the specified domains:
    1. $x^{3}-6x^{2}$ on the domain $[-5,5]$
      Answer: The maximum and minimum occur at one of the local extrema. The candidates for extrema are the endpoints and the critical points. Note that this function $f(x)=x^{3}-6x^{2}$ has derivative $3x^{2}-12x=3x(x-4)$, so it only has two critical points $0$ and $4$ (note that it is differentiable everywhere, so these are the only ones). Now we jut test values: \[f(-5)=-125-6\cdot 5^{2}=-125-150=-275\] \[f(5)=125-150=-25\] \[f(0)=0\] \[f(4)=64-6\cdot 16=-32\] so the maximum value of $f$ on this interval is $0$ and the minimum value is $-275$.
    2. $x^{2}+3x+2$ on $[0,2]$
      Answer: The critical points in this interval is where $f'(x)=2x+3=0$, which is $x=\frac{3}{2}$, so we test this value with the endpoints: \[f(0)=2\] \[f(2)=4+3\cdot 2+2=12\] \[f(\frac{3}{2})=\frac{9}{4}+\frac{9}{2}+3=\frac{39}{4},\] so the maximum value of $f$ in this domain is 12 and the minimum value is 2.
    3. $\ln(x)+\ln (3-x)$ on $[1,2]$. First, let's solve for where the derivative of this function is zero: \[f'(x)=\frac{1}{x}-\frac{1}{3-x}=0\] adding the negative term to both sides gives \[\frac{1}{x}=\frac{1}{3-x}\] and taking one over both sides gives \[x=3-x\] so $x=\frac{3}{2}$ is a solution, and this solution is in our domain. Notice that $0$ and $3$ are also critical points for $f$ since $f'$ does not exist there, but they are not critical points in our domain (they are outside the interval $[1,2]$), so we don't regard them. Testing this value against the endpoints, we get \[f(1)=\ln 2\] \[f(2)=\ln 2\] and \[f(\frac{3}{2})=\ln\frac{3}{2}+\frac{3}{2}=2\ln \frac{3}{2}=\ln\left(\frac{3}{2}\right)^{2}=\ln\frac{9}{4}.\] Note that $\ln $ is an increasing function, so because $\frac{9}{4}>2$, we have $\ln\frac{9}{4}>\ln 2$, hence the maximum value of $f$ in this interval is $\ln\frac{9}{4}$ and the minimum value is $\ln 2$.

     

  4. Find the extrema of the following functions and determine whether they are local maxima or minima:
    1. $x^{2}+4$
      Answer: Setting $f$ equal to this function, the derivative is $f'(x)=2x$, which is zero only when $x=0$. Since $f$ is differentiable everywhere, this is the only critical point. Since $f''(x)=2$, $f''(0)=2>0$, which means $0$ is a local minimum.
    2. $x^{4}+4x^{3}+4x^{2}$
      Answer: $f'(x)=4x^{3}+12x^{2}+8x=4x(x^{2}+3x+2)=4x(x+1)(x+2)$, so this has critical points at $0,-1,-2$. To check which kind of extrema these are, we look at the double derivative \[f''(x)=12x^{2}+24x+8,\] and plugging in our critical points, we get \[f''(0)=8>0,\] \[f''(-1)=12-24+8=-4<0\] \[f''(-2)=48-48+8=8>0\] so $0$ and $-2$ are local minimal and $-1$ is a local maxima.
    3. $\frac{x}{x^{2}+1}$
      Answer: \[f'(x)=\frac{x^{2}+1-x(2x)}{(x^{2}+1)^{2}}=\frac{1-x^{2}}{(x^{2}+1)^{2}}=0\] when $x=\pm 1$, and since $f$ is differentiable everywhere, these are the only critical points. \[f''(x)=\frac{(-2x)(x^{2}+1)^{2}-(1-x^{2})2(x^{2}+1)(2x)}{(x^{2}+1)^{4}}=\frac{-2x(x^{2}+1)-(1-x^{2})4x}{(x^{2}+1)^{3}} =\frac{2x^{3}-6x}{(x^{2}+1)^{3}}.\] Note that the denominator is always positive, so we need only check if the numerator is positive or negative at $\pm 1$ (this simplifies calculations a bit). The numerat or is -4 at $x=1$ and $4$ at $x=-1$, so $x=1$ is a local max and $x=-1$ is a local min.