Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　３．３

　Tｈｅ　Fｉｒｓｔ　Dｅｒｉｖａｔｉｖｅ　Tｅｓｔ　Fｏｒ　Iｎｃｒｅａｓｉｎｇ　ａｎｄ　Dｅｃｒｅａｓｉｎｇ

Ｉｎｔｒｏｄｕｃｔｉｏｎ

• [0:00]. Topic: The First Derivative Test — determining intervals of increase and decrease of a function.
  • Identifying where $𝒇′(𝒙)>0$ or $𝒇′(𝒙)<0$ to locate increasing and decreasing behavior.
  • Application to curve sketching: analyzing the shape of a function by detecting local maxima, minima, and transitions in slope.

Ｒｅｖｉｅｗ　ｏｆ　ｐｒｅｖｉｏｕｓ　ｃｏｎｃｅｐｔｓ

• [0:55]. Review of the meaning of the first derivative and its relationship with a function’s increase and decrease. – [📷image]
  • [1:10]. If $𝒇′(𝒙)>0$, the function $𝒇(𝒙)$ is increasing.
  • [2:16]. If $𝒇′(𝒙)<0$, the function $𝒇(𝒙)$ is decreasing.
  • [2:33]. If $𝒇′(𝒙)=0$, constant, the function $𝒇(𝒙)$ has a horizontal tangent, and a critical number is identified.

Ｔｈｅ　ｆｉｒｓｔ　ｄｅｒｉｖａｔｉｖｅ　ｔｅｓｔ

• [3:05]. Introduction to the First Derivative Test to find relative extrema (maximums and minimums).
• [3:26]. Step-by-step explanation of the First Derivative Test: – [📷image-1] – [📷image-2]
  • ❶ [4:21]. Find the first derivative, $𝒇′(𝒙)$.
  • ❷ [4:40]. Set the first derivative equal to zero, $𝒇′(𝒙)=0$, to find critical numbers.
    • In addition to the values where $𝒇′(𝒙)=0$, you must also include as critical numbers any point where $𝒇′(𝒙)$ does not exist. This is because c̲r̲i̲t̲i̲c̲a̲l̲ ̲n̲u̲m̲b̲e̲r̲s̲ are defined as any 𝒙 in the domain of 𝒇 where either t̲h̲e̲ ̲d̲e̲r̲i̲v̲a̲t̲i̲v̲e̲ ̲ ̲i̲s̲ ̲z̲e̲r̲o̲ ̲ or th̲e̲ ̲d̲e̲r̲i̲v̲a̲t̲i̲v̲e̲ ̲i̲s̲ ̲u̲n̲d̲e̲f̲i̲n̲e̲d̲.. Even if the derivative does not exist at such a point, it can still correspond to a local maximum, a local minimum, or some other important change in the behavior of the function. This kind of points are:
      • Points where the denominator is zero (possible vertical asymptotes).
        • This occurs when the function is not defined (or tends to ±∞) as 𝒙 approaches a certain value 𝒙 = a.
          • Example: $𝒇(𝒙)=\dfrac{1}{𝒙}$. As $𝒙\to 0$, $𝒇(𝒙)\to \pm\infty$, and the line $𝒙=0$ is a vertical asymptote.
      • Sharp corners or cusps (abrupt changes in slope.
      • Vertical tangents (infinite slope).
        • This occurs when the function is defined at a point, but its derivative becomes infinite (or tends to ±∞).
          • Example: $𝒇(𝒙)=𝒙^{1/3}$ at $𝒙=0$. The function is defined at $𝒙=0$, but $𝒇′(𝒙)=\dfrac{1}{3𝒙^{2/3}}$ tends to ±∞ as $𝒙\to 0$, indicating a vertical tangent.
      • Discontinuities (jump, infinite, removable).
  • ❸ [4:58]. Create a first derivative sign chart using the critical numbers.
    • Draw a number line and mark the critical points.
      
      
      
    • Indicate the sign of $𝒇′(𝒙)$ above the number line for each interval.
  • ❹ [6:16]. Test a point in each interval defined by the critical numbers.
    • Determine the sign of $𝒇′(𝒙)$ in each interval.
    • Substitute a sample value into $𝒇′(𝒙)$ to check if the slope is positive or negative.
    • The sign of $𝒇′(𝒙)$ tells whether the function is increasing or decreasing in that interval.
    • Local Maximum:
      • i̲f̲ $𝒇′(𝒙)$ changes from positive to negative at a critical number,
        t̲h̲e̲n̲ there is a local maximum.
    • Local Minimum:
      • i̲f̲ $𝒇′(𝒙)$ changes from negative to positive at a critical number,
        t̲h̲e̲n̲ there is a local minimum.
    • Neither Maximum nor Minimum:
      • i̲f̲ $𝒇′(𝒙)$ does not change sign at a critical number,
        t̲h̲e̲n̲ it is not classified as a maximum or minimum.
      • Sometimes referred to as a horizontal inflection point, if:
        • $𝒇′(𝒙)=0$.
        • The concavity changes (i.e., $𝒇′(𝒙)$ doesn't change sign, but $𝒇″(𝒙)$ does).
        • 🧩 Example –: $𝒇(𝒙)=𝒙^3$
          • $𝒇′(𝒙)=3𝒙^2$ → $𝒇′(0)=0$, but stays positive on both sides of $0$.
          • $𝒇″(𝒙)=6𝒙$ → changes sign at $𝒙=0$.
          • So, $𝒙=0$ is a critical point, not a max or min, but a horizontal inflection point.
• [10:44]. 🧩 Example –: Application of the first derivative test to an example $𝒇(𝒙)=𝒙^3-3𝒙+1$ – [📷image]
  • ❶ Find the first derivative:
    • $𝒇(𝒙)=𝒙^3−3𝒙+1 \Rightarrow 𝒇′(𝒙)=3𝒙^2−3$
  • ❷ Set the first derivative equal to zero:
    • $3𝒙^2−3=0 \Rightarrow 𝒙^2=1 \Rightarrow 𝒙=−1$ and $𝒙=1$
    • There are no points where $𝒇′(𝒙)$ is undefined.
    • Critical numbers: $𝒙=−1$, $𝒙=1$ (both are in the domain of the original funcction)
  • ❸ Check for values that make the derivative undefined.
  • ❸ Create a sign chart for $𝒇′(𝒙)$:
  • ❹ Test a value from each interval in $𝒇′(𝒙)$ to determine the sign.
    • Test values:
      • $𝒙=−2$ → $𝒇′(𝒙)>0$
      • $𝒙=0$ → $𝒇′(𝒙)<0$
      • $𝒙=2$ → $𝒇′(𝒙)>0$
    • Sign chart:
      
      
      
    • Apply the First Derivative Test:
      • At $𝒙=−1$: $𝒇′$ changes from $+$ to $−$ → Local Maximum
      • At $𝒙=1$: $𝒇′$ changes from $−$ to $+$ → Local Minimum
    • Evaluate the function at critical points:
      • $𝒇(−1)=(-1)^3−3(-1)+1=3$
      • $𝒇(1)=1^3−3(1)+1=−1$
  • Answer:
    • Local Maximum at $(−1,3)$
    • Local Minimum at $(1,−1)$
• [17:12]. 🧩 Example –: Presentation of a second example to apply the first derivative test $𝒇(𝒙)=3𝒙^{5/3}-15𝒙^{2/3}$ – [📷image]
  • ❶ Find the first derivative:
    • $𝒇(𝒙)=3𝒙^{5/3}-15𝒙^{2/3} \;\Rightarrow\; 𝒇′(𝒙)=5𝒙^{2/3}-10𝒙^{-1/3}$
  • ❷ Set the first derivative equal to zero: $5𝒙^{2/3}-10𝒙^{-1/3}=0$
    • Multiply both sides by $𝒙^{1/3}$:
      • $5𝒙-10=0 \Rightarrow 𝒙=2$
    • Check where $𝒇′(𝒙)$ is undefined:
      • $𝒇′(𝒙)$ is undefined at $𝒙=0$ (due to $𝒙^{-1/3}$)
    • Critical numbers: $𝒙=0$, $𝒙=2$ (both are in the domain of $𝒇$)
  • ❸ Create a sign chart for $𝒇′(𝒙)$:
    
    
    
  • ❹ Test a value from each interval in $𝒇′(𝒙)$ to determine the sign:
    • Test values:
      • $𝒙=−1 \;\Rightarrow\; 𝒇′(−1)=5(-1)^{2/3}-10(-1)^{-1/3}>0 \;\Rightarrow\; 𝒇′(𝒙)<0$
      • $𝒙=1 \;\Rightarrow\; 𝒇′(1)=5-10<0 \;\Rightarrow\; 𝒇′(𝒙)<0$
      • $𝒙=3 \;\Rightarrow\; 𝒇′(3)=5\cdot 3^{2/3}-10\cdot 3^{-1/3}>0 \;\Rightarrow\; 𝒇′(𝒙)>0$
    • Sign chart:
      
      
      
    • Apply the First Derivative Test:
      • At $𝒙=0$: The derivative $𝒇′$ is undefined due to a vertical tangent, so this point is not considered a local extremum.
        • This will become clearer when we study the second derivative test.
      • At $𝒙=2$: $𝒇′$ changes from $−$ to $+$ ⇒ Local Minimum.
    • Evaluate the function at critical points:
      • $𝒇(0)=3\cdot 0-15\cdot 0=0$
      • $𝒇(2)=3\,(2)^{5/3}-15\,(2)^{2/3}\approx 9.52-23.81\approx -14.29$
  • Answer:
    • Local Minimum at $(2,\,−14.29)$
    • Vertical tangent (not extremum) at $(0,\,0)$