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Combined investigations and families of functions

Combined investigations and families of functions

Two horizontal asymptotes: exex+1\displaystyle \frac{e^x}{e^x+1}

The example below investigates f(x)=exex+1\displaystyle {f(x)}={\frac{e^x}{e^x+1}} – a rational function of ex\displaystyle e^x:

  • The domain: every x\displaystyle x, since ex+1>0\displaystyle e^x+1>0.
  • 0<ex<ex+1\displaystyle 0<e^x<e^x+1, so 0<f(x)<1\displaystyle 0<f(x)<1 for every x\displaystyle x.
  • As x→−∞\displaystyle x\to-\infty, ex→0\displaystyle e^x\to0 and f(x)→0\displaystyle f(x)\to0: the asymptote y=0\displaystyle {y}={0} on the left.
  • As x→∞\displaystyle x\to\infty, f(x)=11+e−x→1\displaystyle {f(x)}={\frac{1}{1+e^{-x}}\to1}: the asymptote y=1\displaystyle {y}={1} on the right.
  • f′(x)=ex(ex+1)−ex⋅ex(ex+1)2=ex(ex+1)2>0\displaystyle {f'(x)}={\frac{e^x(e^x+1)-e^x\cdot e^x}{(e^x+1)^2}}={\frac{e^x}{(e^x+1)^2}>0}: the function increases throughout, with no extremum points.
  • f′′(x)=ex(1−ex)(ex+1)3\displaystyle {f''(x)}={\frac{e^x(1-e^x)}{(e^x+1)^3}}, and its sign is the sign of 1−ex\displaystyle 1-e^x.
f(x)=exex+1\displaystyle f(x)=\frac{e^{x}}{e^{x}+1}
y=0\displaystyle y=0
y=1\displaystyle y=1
P=(0,0.5)\displaystyle P=(0,0.5)
  • Where x<0\displaystyle x<0, ex<1\displaystyle e^x<1 and f′′(x)>0\displaystyle f''(x)>0 – concave up; where x>0\displaystyle x>0, f′′(x)<0\displaystyle f''(x)<0 – concave down.
  • An inflection point: (0,12)\displaystyle \left(0,\frac{1}{2}\right) – where the function increases fastest.
A vertical tangent at the end of the domain: ex−1\displaystyle \sqrt{e^x-1}

The example below investigates f(x)=ex−1\displaystyle {f(x)}={\sqrt{e^x-1}}:

  • The domain: ex−1≥0\displaystyle e^x-1\geq0, that is ex≥1\displaystyle e^x\geq1 and x≥0\displaystyle x\geq0.
  • f(0)=0\displaystyle {f(0)}={0}: the graph starts at the origin.
  • f′(x)=ex2ex−1>0\displaystyle {f'(x)}={\frac{e^x}{2\sqrt{e^x-1}}>0} where x>0\displaystyle x>0: the function increases throughout, and the end point (0,0)\displaystyle (0,0) is a minimum point.
  • As x→0+\displaystyle x\to0^+, the denominator tends to 0 and the numerator to 1, so f′(x)→∞\displaystyle f'(x)\to\infty: the graph leaves the origin upright, and the tangent there is the y-axis.
  • As x→∞\displaystyle x\to\infty, f(x)→∞\displaystyle f(x)\to\infty: no horizontal asymptote.
f(x)=ex−1\displaystyle f(x)=\sqrt{e^{x}-1}
Investigating ln⁡xx\displaystyle \frac{\ln x}{\sqrt{x}}

The example below investigates f(x)=ln⁡xx\displaystyle {f(x)}={\frac{\ln x}{\sqrt{x}}}:

  • The domain: x>0\displaystyle x>0. The x-axis: (1,0)\displaystyle (1,0).
  • As x→0+\displaystyle x\to0^+, ln⁡x→−∞\displaystyle \ln x\to-\infty and x→0+\displaystyle \sqrt{x}\to0^+, so f(x)→−∞\displaystyle f(x)\to-\infty: the vertical asymptote x=0\displaystyle {x}={0}.
  • As x→∞\displaystyle x\to\infty, ln⁡x\displaystyle \ln x grows more slowly than x\displaystyle \sqrt{x}, so f(x)→0\displaystyle f(x)\to0: the horizontal asymptote y=0\displaystyle {y}={0}.
  • f′(x)=1x⋅x−ln⁡x⋅12xx=2−ln⁡x2xx\displaystyle {f'(x)}={\frac{\frac{1}{x}\cdot\sqrt{x}-\ln x\cdot\frac{1}{2\sqrt{x}}}{x}}={\frac{2-\ln x}{2x\sqrt{x}}}, and f′(x)=0\displaystyle {f'(x)}={0} where x=e2\displaystyle {x}={e^2}.
  • f\displaystyle f increases where 0<x<e2\displaystyle 0<x<e^2 and decreases where x>e2\displaystyle x>e^2: a maximum (e2,2e)≈(7.39,0.74)\displaystyle \left(e^2,\frac{2}{e}\right)\approx(7.39,0.74).
f(x)=ln⁡xx\displaystyle f(x)=\frac{\ln x}{\sqrt{x}}
x=0\displaystyle x=0
M=(7.389,0.736)\displaystyle M=(7.389,0.736)
A family with a parameter: x2−aln⁡x\displaystyle x^2-a\ln x

The example below investigates the family f(x)=x2−aln⁡x\displaystyle {f(x)}={x^2-a\ln x}, with a>0\displaystyle a>0 a parameter:

  • The domain: x>0\displaystyle x>0, for every a\displaystyle a.
  • f′(x)=2x−ax=2x2−ax\displaystyle {f'(x)}={2x-\frac{a}{x}}={\frac{2x^2-a}{x}}, and f′(x)=0\displaystyle {f'(x)}={0} where x=a2\displaystyle {x}={\sqrt{\frac{a}{2}}} (the negative solution is not in the domain).
  • f′′(x)=2+ax2>0\displaystyle {f''(x)}={2+\frac{a}{x^2}>0}: every function of the family is concave up, so the point is a minimum and there are no inflection points.

On the graph: a=2\displaystyle {a}={2} and a=8\displaystyle {a}={8}.

f(x)=x2−2ln⁡x\displaystyle f(x)=x^{2}-2\ln x
g(x)=x2−8ln⁡x\displaystyle g(x)=x^{2}-8\ln x
  • a=2\displaystyle {a}={2}: the minimum (1,1)\displaystyle (1,1).
  • a=8\displaystyle {a}={8}: the minimum (2,4−8ln⁡2)≈(2,−1.55)\displaystyle (2,4-8\ln2)\approx(2,-1.55).
  • The larger a\displaystyle a, the further right the minimum point: the parameter changes where the extremum is, not its type.
Exercise 1

The function f(x)=exx2\displaystyle {f(x)}={\frac{e^x}{x^2}} is given, for x≠0\displaystyle x\neq0.

a

Find f′(x)\displaystyle f'(x).

f(x)=exx2\displaystyle f(x)=\frac{e^{x}}{x^{2}}

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b

    Solve the equation f′(x)=0\displaystyle {f'(x)}={0}.

    c

    Mark the extremum point on the graph.
    Add a row for each of the two asymptotes: the vertical one, and the horizontal one on the left.

    f(x)=exx2\displaystyle f(x)=\frac{e^{x}}{x^{2}}

    Exercise 2

    The function f(x)=xln⁡x\displaystyle {f(x)}={\sqrt{x}\ln x} is given, for x>0\displaystyle x>0.

    a

    Find f′(x)\displaystyle f'(x).

    f(x)=xln⁡x\displaystyle f(x)=\sqrt{x}\ln x

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

      b

      Solve the equation f′(x)=0\displaystyle {f'(x)}={0}.

      c

      Find f′′(x)\displaystyle f''(x).

      f(x)=xln⁡x\displaystyle f(x)=\sqrt{x}\ln x

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        d

        Find where the function is concave up: solve the inequality f′′(x)>0\displaystyle f''(x)>0.

        e

        Mark the extremum point and the inflection point on the graph.

        f(x)=xln⁡x\displaystyle f(x)=\sqrt{x}\ln x

        Exercise 3

        The family f(x)=ex−ax\displaystyle {f(x)}={e^x-ax} is given, with a>0\displaystyle a>0 a parameter.

        a

        Find f′(x)\displaystyle f'(x).

        f(x)=ex−ax\displaystyle f(x)=e^{x}-ax

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

          b

          Show that for every a>0\displaystyle a>0 the function has one extremum point, where x=ln⁡a\displaystyle {x}={\ln a}, and find its type.

          c

          Find the y-coordinate of the extremum point.

          f(x)=ex−ax\displaystyle f(x)=e^{x}-ax

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

            d

            The extremum point is on the x-axis.
            Find a\displaystyle a.

              e

              Substitute the a\displaystyle a you found, and mark the extremum point on the graph.

              f(x)=ex−ex\displaystyle f(x)=e^{x}-ex

              Exercise 4

              The function f(x)=xln⁡x\displaystyle {f(x)}={\frac{x}{\ln x}} is given.

              a

              Find the domain, and explain.

              b

              Find f′(x)\displaystyle f'(x).

              f(x)=xln⁡x\displaystyle f(x)=\frac{x}{\ln x}

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                c

                Solve the equation f′(x)=0\displaystyle {f'(x)}={0}.

                d

                Mark the extremum point on the graph, and add a row with the equation of the vertical asymptote.

                f(x)=xln⁡x\displaystyle f(x)=\frac{x}{\ln x}