MATH 1830 Notes

Unit 3 Applications of Derivatives

3.3 Second Derivative Test

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3.2 HW

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Second Derivative Test

Analyze $f''(x)$ to Identify Intervals of Concavity and Points of Inflection on the Graph of $f(x).$

  1. Find $f''(x).$
  2. Find the critical numbers for the function.
    1. Values of $x$ where $f''(x)=0$ are critical numbers.
    2. Values of $x$ where $f''(x)$ is undefined are partitions.
    3. Values of $x$ where $f(x)$ is undefined are partitions.
  3. Graph the critical numbers and partitions on a number line, separating the number line into intervals.
  4. Determine the intervals on which $f(x)$ is concave up or concave down.
    1. Test one point contained in the interval (do not use the end points of the interval).
    2. $f(x)$ is concave down on the interval if $f''(x)<0$.
    3. $f(x)$ is concave up on the interval if $f''(x)>0$.
  5. Identify inflection points of $f(x)$. A point of inflection occurs at $x=a$ when $f''(a)=0$ and $f''(x)$ changes concavity across $a$.

Use the Second Derivative Test to analyze the function. Identify intervals of concavity and points of inflection on the graph of the function.

  1. $f(x)={{x}^{3}}+6{{x}^{2}}+9x$

    1. Concave Up and Concave Down

      ${f}''(x)=6x+12$

      Values of x where ${f}''(x)=0:$

      $6x+12=0$

      $6x=-12$

      $x=-2$

      Values of x where ${f}''\left( x \right)$ is undefined:

      There are no values of x where ${f}''\left( x \right)$ is undefined.

      Values of x where $f\left( x \right)$ is undefined:

      There are no values of x where $f\left( x \right)$ is undefined.

      Separate into intervals using: $x=-2$.

      Graph of Second Derivative of f of x and how it relates to the x axis. Intervals are separated by values of x where the second derivative equals 0 or is undefined. Values where f of x is undefined also serve to segment intervals. The graph indicates intervals where the second derivative of f of x is positive and where the second derivative is negative.
      Sign chart for ${f}''\left( x \right)$: negative, ${f}''\left( -2 \right)=0$, positive

      Concave up:

      The graph of $f(x)$ is concave up on the interval $\left( -2,\infty \right)$.

      Concave down:

      The graph of $f(x)$ is concave down on the interval $\left( -\infty ,-2 \right)$.

    2. Inflection Points:

      $f\left( -2 \right)={{\left( -2 \right)}^{3}}+6{{\left( -2 \right)}^{2}}+9\left( -2 \right)=-2$.

      The graph of $f(x)$ has a point of inflection at $\left( -2,-2 \right)$

      blank 4 quadrant coordinate plane

      Graph of F of x indicating intervals of increase, intervals of decrease, local maxima and local minima

  2. $f(x)=\frac{3x+4}{2x-5}$

    1. Concave Up and Concave Down

      ${f}''\left( x \right)=-23\left( -2 \right){{\left( 2x-5 \right)}^{-3}}\left( 2 \right)=\frac{92}{{{\left( 2x-5 \right)}^{3}}}$

      Values of x where ${f}''(x)=0$:

      There are no values of x where ${f}''\left( x \right)=0$.

      Values of x where ${f}''\left( x \right)$ is undefined:

      $ 2x-5=0$

      $ x=\frac{5}{2}$

      Values of x where $f\left( x \right)$is undefined:

      $2x-5=0$

      $x=\frac{5}{2}$

      Separate into intervals using: $x=\frac{5}{2}$.

      Graph of Second Derivative of f of x and how it relates to the x axis. Intervals are separated by values of x where the second derivative equals 0 or is undefined. Values where f of x is undefined also serve to segment intervals. The graph indicates intervals where the second derivative of f of x is positive and where the second derivative is negative.
      Sign chart for ${f}''\left( x \right)$: negative, ${f}''\left( \frac{5}{2} \right)=$ undefined, positive

      Concave up:

      The graph of $f(x)$ is concave up on the interval $\left( \frac{5}{2},\infty \right).$

      Concave down:

      The graph of $f(x)$ is concave down on the interval $\left( -\infty ,\frac{5}{2} \right).$

    2. Inflection Points:

      Even though the function changes from concave down to concave up at $x=\frac{5}{2}$, there are no points of inflection because the function is undefined at $x=\frac{5}{2}.$

      blank 4 quadrant coordinate plane

      Graph of f of x indicating intervals of increase, intervals of decrease, local maxima and local minima

  3. The annual first quarter change in revenue for Apple, Inc. is given by the regression model: $$f(x)=-0.005x^4+0.113x^3-0.889x^2+7.946x-5.346$$ where x is Years Since 1998.

    Use the Second Derivative Test to analyze the function. Identify intervals of concavity and points of inflection on the graph of the function.

    $f''(x)=-0.06x^2+0.678x-1.778$

    1. Identify Critical Values and Partitions for the Sign Chart.

      Values of x where $f''(x)=0:$

      $f''(x)=0$ at $x= 4.137$ and $x= 7.163$

      Values of x where $f''(x)$ is undefined:

      There are no values of x where $f''(x)$ is undefined.

      Values of x where $f(x)$ is undefined:

      There are no values of x where $f(x)$ is undefined.

      Separate into intervals using: $x=4.137$ and $x=7.163$.

      graph of second derivative
      Sign graph of $f''(x)$ reading left to right: negative, ${f}'\left(4.137\right)=0,$ positive, ${f}'\left(7.163\right)=0,$ negative
      Intervals of Concavity

      Concave Up:

      $(4.137, 7.163)$

      The percent change in revenue is increasing at an increasing rate from $x=4.137$ to $x=7.163$.

      Concave down:

      $\lbrack0,\;4.137)\;\cup\;(7.163,\;18\rbrack$

      The percent change in revenue is increasing at a decreasing rate from $x=0$ to $x=4.137$.

      The percent change in revenue is increasing at a decreasing rate from $x=7.163$ to $x=12.358$ (the maximum) and decreasing at an increasing rate from $x=12.358$ to $x=18$.

    2. Inflection points:

      $(4.137, 18.848)$ and $(7.163, 34.325)$

      According to the model, the percent change in revenue is changing most rapidly at $x=4.137$ and $x=7.163$.

    3. Graph $f(x).$

      blank coordinate plane

      graph of revenue function

  4. Using data from the Federal Reserve, the Dow S&P 500 annual percent return on investments for the years 2008-2014 can be modeled by the following equation: $$A(t) = - 1.64{t^4} + 20.85{t^3} - 86.05{t^2} + 127.87t - 36.24$$ where t is in years since 2008 and A(t) is in percent.

    Source: http://pages.stern.nyu.edu/~adamodar/New_Home_Page/datafile/histretSP.html

    Use the Second Derivative Test to analyze the function. Identify intervals of concavity and points of inflection on the graph of the function.

    ${A}''\left( t \right)=-19.68{{t}^{2}}+125.1t-172.10$

    1. Graph the critical numbers on a number line and determine the sign for each interval.

      Values of t where ${A}''(t)=0.$

      $t \approx 2.013$ and $t \approx 4.343$

      Values of t where ${A}''\left( t \right)$ is undefined.

      There are no values of t where ${A}''(t)$ is undefined.

      Values of t where $A\left( t \right)$ is undefined.

      There are no values of t where $A\left( t \right)$ is undefined.

      Separate into intervals using: $t=2.013$ and $t=4.343$.

      Graph of the second derivative of the function A of t
      Sign chart for ${A}''\left( t \right)$ reading left to right: negative, ${A}''\left( 2.013 \right)\approx 0$ , positive, ${A}''\left( 4.343 \right)\approx 0$, negative

    2. Determine the intervals on which $A(t)$ is concave up / concave down.

      On what intervals is A(t) concave up? Interpret these results.

      The function is concave up on the interval: $\left( 2.013,4.343 \right).$

      On the interval $\left( 2.01,3.10 \right),$ the annual percent return on investments is decreasing at a decreasing rate each year.

      On the interval $\left( 3.10,4.34 \right),$ the annual percent return on investments is increasing at an increasing rate each year.

      On what intervals is $A(t)$ concave down? Interpret these results.

      The function is concave down on the intervals: $\left( 0,2.013 \right)\cup \left( 4.343,6 \right).$

      On the intervals from $\left( 0,1.2 \right)\,\text{and}\,\left( 4.34,5.23 \right),$ the annual percent return on investments is increasing at a decreasing rate each year.

      On the interval $\left( 3.10,4.34 \right),$ the annual percent return on investments is decreasing at an increasing rate each year.

    3. Identify points of inflection for $A(t)$. Interpret these results.

      There are two points of inflection:

      $\left( 2.013,15.618 \right)$ and $\left( 4.343,20.557 \right)$

      According to the model, the annual percent return on investments was changing most rapidly at $t \approx 2.01$ and $t \approx 4.34$ (in years after 2008).

    4. Graph $A(t).$

      blank coordinate plane

      Graph of A of t indicating intervals of increase, intervals of decrease, local maxima and local minima, x axis: 0 to 6 years after 2008, y axis: -40 to 40 percent

3.3 Second Derivative Test Homework

Use the Second Derivative Test to analyze the function. Identify intervals of concavity and points of inflection on the graph of the function.

  1. $f(x)=3{{x}^{2}}+5x-2$

    1. Concave Up and Concave Down

      ${f}''\left( x \right)=6$

      Values of x where ${f}''(x)=0:$

      There are no values of x where ${f}''\left( x \right)=0.$

      Values of x where ${f}''\left( x \right)$ is undefined:

      There are no values of x where ${f}''\left( x \right)$ is undefined.

      Values of x where $f\left( x \right)$ is undefined:

      There are no values of x where $f\left( x \right)$ is undefined.

      Because there are no x values that meet the requirements for a partition or critical value, $f(x)$ will maintain the same concavity across the entire domain.

      Graph of Second Derivative of f of x and how it relates to the x axis. Intervals are separated by values of x where the second derivative equals 0 or is undefined. Values where f of x is undefined also serve to segment intervals. The graph indicates intervals where the second derivative of f of x is positive and where the second derivative is negative.
      Sign chart for ${f}''\left( x \right)$: positive on domain

      Concave up:

      The graph of $f(x)$ is concave up on the interval $\left( -\infty ,\infty \right)$.

      Concave down:

      There are no intervals where $f(x)$ is concave down.

    2. Inflection Points:

      There are no points of inflection.

      blank 4 quadrant coordinate plane

      Graph of F of x indicating intervals of increase, intervals of decrease, local maxima and local minima

  2. $f(x)=\frac{4x-7}{5x+1}$

    1. Concave Up and Concave Down

      ${f}''\left( x \right)=39\left( -2 \right){{\left( 5x+1 \right)}^{-3}}\left( 5 \right)=-\frac{390}{{{\left( 5x+1 \right)}^{3}}}$

      Values of x where ${f}''(x)=0$:

      There are no values of x where ${f}''\left( x \right)=0$.

      Values of x where ${f}''\left( x \right)$ is undefined:

      $ 5x+1=0$

      $ x=-\frac{1}{5}$

      Values of x where $f\left( x \right)$is undefined:

      $5x+1=0$

      $x=-\frac15$

      Separate into intervals using: $x=-\frac{1}{5}$.

      Graph of Second Derivative of f of x and how it relates to the x axis. Intervals are separated by values of x where the second derivative equals 0 or is undefined. Values where f of x is undefined also serve to segment intervals. The graph indicates intervals where the second derivative of f of x is positive and where the second derivative is negative.
      Sign chart for ${f}''\left( x \right)$: positive, ${f}''\left( -\frac{1}{5} \right)=$ undefined, negative

      Concave up:

      The graph of $f(x)$ is concave up on the interval $\left( -\infty ,-\frac{1}{5} \right).$

      Concave down:

      The graph of $f(x)$ is concave down on the interval $\left( -\frac{1}{5},\infty \right).$

    2. Inflection Points:

      Even though the function changes from concave up to concave down at $x=-\frac{1}{5}$, there are no points of inflection because the function is undefined at $x=-\frac{1}{5}.$

      blank 4 quadrant coordinate plane

      Graph of f of x indicating intervals of increase, intervals of decrease, local maxima and local minima

  3. Using data from Statista, the total annual amount spent on the purchase of golf equipment in the United States for the years 2008-2014 can be modeled by the following equation:

    $$A(t)=-31.94{{t}^{3}}+301.16{{t}^{2}}-665.61t+3454.63$$ where t is in years since 2008 and A(t) is in millions of dollars.

    Source: href="https://www.statista.com/statistics/201038/purchases-of-golf-equipment-in-the-us-since-2007/"

    Use the Second Derivative Test to analyze the function. Identify intervals of concavity and points of inflection on the graph of the function.

    Analyze ${A}''\left( t \right)$

    ${A}''\left( t \right)=-191.64t+602.32$

    1. Graph the critical numbers on a number line and determine the sign for each interval.

      Values of t where ${A}''(t)=0$

      $t\approx 3.14$ years since 2008

      Values of t where ${A}''\left( t \right)$ is undefined.

      There are no values of t where ${A}''(t)$ is undefined

      Values of t where $A\left( t \right)$ is undefined.

      There are no values of t where $A\left( t \right)$ is undefined

      Separate into intervals using: $t=3.14$.

      Graph of the second derivative of the function A of t

      Sign chart for ${A}''\left( t \right)$ reading left to right: positive, ${A}''\left( 3.14 \right)\approx 0$, negative

    2. Determine the intervals on which $A(t)$ is concave up / concave down

      On what intervals is $A(t)$ concave up? Interpret these results.

      The function is concave up on the interval: $\left( 0,3.14 \right)$

      When interpreting concavity results, we will segment the interval $\left( 0,3.14 \right)$ into sections based on increasing/decreasing behavior of $f(x)$.

      On the interval from $\left( 0,1.43 \right)$ the amount spent annually was decreasing at a decreasing rate. On the interval from $\left( 1.43,3.14 \right)$, the amount spent each year was increasing at an increasing rate.

      On what intervals is $A(t)$ concave down? Interpret these results.

      The function is concave down on the intervals: $\left( 3.14,6 \right)$

      On the interval from $\left( 3.14,4.86 \right)$ the amount spent annually is increasing at a decreasing rate. From $\left( 4.86,6 \right)$ the amount spent each year was decreasing at an increasing rate.

      Identify points of inflection for A(t). Interpret these results.

      There is a point of inflection at $t=3.14$ years since 2008.

      $A\left( 3.14 \right)=-31.94{{\left( 3.14 \right)}^{3}}+301.16{{\left( 3.14 \right)}^{2}}-665.61\left( 3.14 \right)+3454.63\approx 3345.1$

      The actual rate of change on spending at $t=3.14$ years since 2008 is:

      ${A}'\left( 3.14 \right)=-95.82{{\left( 3.14 \right)}^{2}}+602.32\left( 3.14 \right)-665.61\approx 280.93$

      According to the model, the amount of money spent on golf equipment was changing most rapidly at x=3.14 (in years after 2008). At that time, the annual amount spent was increasing at a rate of \$280.93 million per year and the total amount spent was $3345.1 million.

    3. Graph $A(t)$

      Clearly marking the characteristics identified above. Use graphing paper and colored pencils. Your graph should cover an entire piece of graph paper.

      x axis: Years Since 2008 on interval 0 to 6, inclusive. y axis: Purchase of Golf Equipment in Millions of Dollars with range 0 to 5000.

    4. In order to accurately use the model, t values should be interpreted discretely, since the data used was given in discrete values (2008, 2009, etc) Using only discrete values for years, identify the relative maximums and minimums from the function:

      Bar graph showing total amount spent in the United States on golf equipment from 2007 to 2015. x-axis: years 2007-2015. y-axis: total purchases in millions of dollars from 0 to 5000

Additional Resources