Skip to main content

Subsection 3.6.2 First Derivative — Increasing or Decreasing

Now we move on to the first derivative, \(f'(x)\text{.}\) This is a good time to revisit the mean-value theorem (Theorem 2.13.5) and some of its consequences (Corollary 2.13.12). There we considered any function \(f(x)\) that is continuous on an interval \(A\le x\le B\) and differentiable on \(A\lt x\lt B\text{.}\) Then

  • if \(f'(x) \gt 0\) for all \(A \lt x \lt B\text{,}\) then \(f(x)\) is increasing on \((A,B)\)

    — that is, for all \(A \lt a \lt b \lt B\text{,}\) \(f(a) \lt f(b)\text{.}\)

  • if \(f'(x) \lt 0\) for all \(A \lt x \lt B\text{,}\) then \(f(x)\) is decreasing on \((A,B)\)

    — that is, for all \(A \lt a \lt b \lt B\text{,}\) \(f(a) \gt f(b)\text{.}\)

Thus the sign of the derivative indicates to us whether the function is increasing or decreasing. Further, as we discussed in Section 3.5.1, we should also examine points at which the derivative is zero — critical points — and points where the derivative does not exist. These points may indicate a local maximum or minimum.

We will now consider a function \(f(x)\) that is defined on an interval \(I\text{,}\) except possibly at finitely many points of \(I\text{.}\) If \(f\) or its derivative \(f'\) is not defined at a point \(a\) of \(I\text{,}\) then we call \(a\) a singular point 1 This is the extension of the definition of “singular point” mentioned in the footnote in Definition 3.5.6. of \(f\text{.}\)

After studying the function \(f(x)\) as described above, we should compute its derivative \(f'(x)\text{.}\)

  • Critical points — determine where \(f'(x)=0\text{.}\) At a critical point, \(f\) has a horizontal tangent.
  • Singular points — determine where \(f'(x)\) is not defined. If \(f'(x)\) approaches \(\pm\infty\) as \(x\) approaches a singular point \(a\text{,}\) then \(f\) has a vertical tangent there when \(f\) approaches a finite value as \(x\) approaches \(a\) (or possibly approaches \(a\) from one side) and a vertical asymptote when \(f(x)\) approaches \(\pm\infty\) as \(x\) approaches \(a\) (or possibly approaches \(a\) from one side).
  • Increasing and decreasing — where is the derivative positive and where is it negative. Notice that in order for the derivative to change sign, it must either pass through zero (a critical point) or have a singular point. Thus neighbouring regions of increase and decrease will be separated by critical and singular points.

Consider the function

\begin{align*} f(x) &= x^4-6x^3 \end{align*}

  • Before we move on to derivatives, let us first examine the function itself as we did above.

    • As \(f(x)\) is a polynomial its domain is all real numbers.
    • Its \(y\)-intercept is at \((0,0)\text{.}\) We find its \(x\)-intercepts by factoring
      \begin{align*} f(x) &=x^4-6x^3 = x^3(x-6) \end{align*}
      So it crosses the \(x\)-axis at \(x=0,6\text{.}\)
    • Again, since the function is a polynomial it does not have any vertical asymptotes. And since
      \begin{align*} \lim_{x \to \pm \infty} f(x) &= \lim_{x \to \pm \infty} x^4(1-6/x) = +\infty \end{align*}
      it does not have horizontal asymptotes — it blows up to \(+\infty\) as \(x\) goes to \(\pm\infty\text{.}\)

    • We can also determine where the function is positive or negative since we know it is continuous everywhere and zero at \(x=0,6\text{.}\) Thus we must examine the intervals

      \begin{align*} (-\infty,0)&& (0,6) && (6,\infty) \end{align*}

      When \(x \lt 0\text{,}\) \(x^3 \lt 0\) and \(x-6 \lt 0\) so \(f(x) = x^3(x-6) = (\text{negative})(\text{negative}) \gt 0\text{.}\) Similarly when \(x \gt 6\text{,}\) \(x^3 \gt 0, x-6 \gt 0\) we must have \(f(x) \gt 0\text{.}\) Finally when \(0 \lt x \lt 6\text{,}\) \(x^3 \gt 0\) but \(x-6 \lt 0\) so \(f(x) \lt 0\text{.}\) Thus

      interval \((-\infty,0)\) 0 \((0,6)\) 6 \((6,\infty)\)
      \(f(x)\) positive 0 negative 0 positive
    • Based on this information we can already construct a rough sketch.
  • Now we compute its derivative
    \begin{align*} f'(x) &= 4x^3-18x^2 = 2x^2(2x-9) \end{align*}

  • Since the function is a polynomial, it does not have any singular points, but it does have two critical points at \(x=0, 9/2\text{.}\) These two critical points split the real line into 3 open intervals

    \begin{align*} (-\infty, 0) && (0,9/2) && (9/2,\infty) \end{align*}

    We need to determine the sign of the derivative in each intervals.

    • When \(x \lt 0\text{,}\) \(x^2 \gt 0\) but \((2x-9) \lt 0\text{,}\) so \(f'(x) \lt 0\) and the function is decreasing.
    • When \(0 \lt x \lt 9/2\text{,}\) \(x^2 \gt 0\) but \((2x-9) \lt 0\text{,}\) so \(f'(x) \lt 0\) and the function is still decreasing.
    • When \(x \gt 9/2\text{,}\) \(x^2 \gt 0\) and \((2x-9) \gt 0\text{,}\) so \(f'(x) \gt 0\) and the function is increasing.

    We can then summarise this in the following table

    interval \((-\infty,0)\) 0 \((0,9/2)\) 9/2 \((9/2,\infty)\)
    \(f'(x)\) negative 0 negative 0 positive
    decreasing horizontal
    tangent    		 decreasing minimum increasing

    Since the derivative changes sign from negative to positive at the critical point \(x=9/2\text{,}\) this point is a minimum. Its \(y\)-value is

    \begin{align*} y&=f(9/2) = \frac{9^3}{2^3}\left(\frac{9}{2} - 6\right)\\ &= \frac{3^6}{2^3} \cdot \left(\frac{-3}{2} \right) = -\frac{3^7}{2^4} \end{align*}

    On the other hand, at \(x=0\) the derivative does not change sign; while this point has a horizontal tangent line it is not a minimum or maximum.

  • Putting this information together we arrive at a quite reasonable sketch.

    To improve upon this further we will examine the second derivative.

login