Stationary point, Increasing and Decreasing Function, and Curve Sketching
Stationary point
\begin{equation*} f'(x) = 0 \end{equation*}
Nature of stationary point
\begin{equation*} \begin{split} f”(x) & > 0 \quad \text{local minimum} \\\\ f”(x) & < 0 \quad \text{local maximum} \end{split} \end{equation*}
Increasing function
\(f’ (x)>0\)
Decreasing function
\(f’ (x)<0\)
Curve sketching
- Determine curve intercept with axis
- Determine stationary points (if exists)
- Determine nature of stationary points (if exists)
Example
Given a function \(f(x) = x^4 – 2 x^3 + x^2\)
A. Determine its stationary points and its nature
B. Find the inverval where the function is increasing and decreasing
C. Sketch the curve
A. Stationary points
\begin{equation*} \begin{split} & f’ (x) = 0 \\\\ & 4 x^{3} – 6 x^{2} + 2 x = 0 \\\\ & 2 x \:.\: (2x^{2} – 3x + 1) = 0\\\\ & 2x \:.\: (2x – 1)(x – 1) = 0\\\\ & x = 0 \text{ or } x = \frac{1}{2} \text{ or } x = 1 \end{split} \end{equation*}
Nature of stationary points
\begin{equation*} \begin{split} & f” (x) = 12 x^{2} – 12 x + 2 \\\\ & f” (0) = 2 \quad {\color {blue} \text{(local minimum)}} \\\\ & f” (\frac 12) = -1 \quad {\color {blue} \text{(local maximum)}} \\\\ & f” (1) = 2 \quad {\color {blue} \text{(local minimum)}} \end{split} \end{equation*}
Coordinate of stationary points
\begin{equation*} \begin{split} & f(x) = x^4 – 2 x^3 + x^2 \\\\ & f(0) = 0 \\\\ & f \left(\frac 12 \right) = \frac {1}{16} \\\\ & f(1) = 0 \end{split} \end{equation*}
Coordinate of stationary points are:
\((0,0)\) as local minimum
\(\left(\dfrac 12,\dfrac {1}{16} \right)\) as local maximum
\((1,0)\) as local minimum
B. Increasing and decreasing interval
Increasing function at the interval \(0 < x < \frac 12 \) dan \(x > 1\)
Decreasing function at the interval \(x < 0\) dan \(\frac 12 < x < 1\)
C. Curve sketching
Exercise
[s2If current_user_can(access_s2member_level1) and current_user_can(access_s2member_ccap_a_math)]
Determine the interval which the curve \(f(x)=x^{3}+9x^{2}+15x+5\) is increasing and decreasing.
Determine the interval which the curve \(f(x)=x^{3}+9x^{2}+15x+5\) is increasing and decreasing.
Solution:
First derivative
\begin{equation*} \begin{split} & f’ (x) = 0\\\\ & 3x^{2}+18x+15 = 0 \\\\ & x^{2}+6x+5 = 0\\\\ & (x+5)(x+1) = 0\\\\ & x = -5 \text{ or } x = -1 \end{split} \end{equation*}
Increasing function
\begin{equation*} \begin{split} & f'(x) > 0 \\\\ & x<-5 \text{ atau } x>-1 \end{split} \end{equation*}
Decreasing function
\begin{equation*} \begin{split} & f'(x) < 0 \\\\ & -5 < x < 1 \end{split} \end{equation*}
Determine the values of \(k\) which the curve \(f (x) = 3x^3 + (k + 1) x^2 + (k – 1) x + 11\) is increasing.
Determine the values of \(k\) which the curve \(f (x) = 3x^3 + (k + 1) x^2 + (k – 1) x + 11\) is increasing.
Solution:
\begin{equation*} \begin{split} & f'(x) > 0\\\\ & 9x^{2} + 2(k + 1) x + (k – 1) > 0\\\\ & a = 9 \quad b = 2(k + 1) \quad c = (k – 1) \end{split} \end{equation*}
Positive definite
\begin{equation*} \begin{split} & D < 0\\\\ & b^{2} – 4ac < 0\\\\ & [2(k + 1)]^{2} – 4 \:.\: 9 \:.\: (k – 1) < 0\\\\ & 4k^{2} + 8k + 4 – 36k + 36 < 0\\\\ & 4k^{2} – 28k + 40 < 0 \\\\ & k^{2} – 7k + 10 < 0\\\\ & (k – 2)(k – 5) < 0 \end{split} \end{equation*}
\(2 < k < 5\)
Determine the maximum and minimum values of the function \(f(x) = -x^{3} + 6x^{2} + 15x – 2\) for \(-2 < x < 4\).
Determine the maximum and minimum values of the function \(f(x) = -x^{3} + 6x^{2} + 15x – 2\) for \(-2 < x < 4\).
Solution:
Stationary points
\begin{equation*} \begin{split} & f’ (x) = 0\\\\ & -3x^{2}+ 12x + 15 = 0 \\\\ & x^{2} – 4x – 5 = 0\\\\ & (x + 1)(x – 5) = 0\\\\ & x = – 1 \text{ or } x = 5 \end{split} \end{equation*}
\(x = -1\) is in the interval \(-2 < x < 6\)
\(x = 5\) is not in the interval \(-2 < x < 6\), then \(x = 5\) cannot be used.
\begin{equation*} \begin{split} & f(x) = -x^{3} + 6x^{2} + 15x – 2 \\\\ & f (-1) = – (-1)^{3} + 6 (-1)^{2} + 15 (-1) -2 = -10 \quad \color {blue} { \text{(minimum)}} \\\\ & f (-2) = -(-2)^{3} + 6 (-2)^{2} + 15 (-2) – 2 = 0 \\\\ & f (4) = -(4)^{3} + 6 (4)^{2} + 15 (4) – 2 = 88 \quad \color {blue} { \text{(maximum)}} \end{split} \end{equation*}
Maximum value is \(88\)
Minimum value is \(-10\)
The curve \(y=x^{4}- 8x^{3} + 18x^{2} + m\) in the interval \(-1 ≤ x ≤ 2\) has maximum value of 30.
Find the value of m.
The curve \(y=x^{4}- 8x^{3} + 18x^{2} + m\) in the interval \(-1 ≤ x ≤ 2\) has maximum value of 30.
Find the value of m.
Solution:
Stationary points
\begin{equation*} \begin{split} & f’ (x) = 0\\\\ & 4x^{3} – 24x^{2} + 36x = 0 \\\\ & x^{3} – 6x^{2} + 9x = 0\\\\ & x (x-3)^{2} = 0\\\\ & x = 0 \text{ atau } x = 3 \end{split} \end{equation*}
\(x = 0\) is in the interval \(-1 ≤ x ≤ 2\)
\(x = 3\) is not in the interval \(-1 ≤ x ≤ 2 \), then \(x = 3\) cannot be used.
\begin{equation*} \begin{split} & y = x^{4}- 8x^{3} + 18x^{2} + m \\\\ & f (0) = (0)^{4} – 8 (0)^{3} + 18 (0)^{2} + m = m\\\\ & f (-1) = (-1)^{4} – 8 (-1)^{3 }+ 18 (-1)^{2} + m = m + 27\\\\ & f (2) = (2)^{4} – 8 (2)^{3} + 18 (2)^{2} + m = m + 24 \end{split} \end{equation*}
Maximum value is \((m + 27)\).
\begin{equation*} \begin{split} & m + 27 = 30 \\\\ & m = 3 \end{split} \end{equation*}
The curve \(y = x^{3} + ax^{2} + bx + c\) have coordinates of minimum point of \((3, -69)\) and maximum point of \((-5,m)\).
Find the value of m.
The curve \(y = x^{3} + ax^{2} + bx + c\) have coordinates of minimum point of \((3, -69)\) and maximum point of \((-5,m)\).
Find the value of m.
Solution:
Stationary points
\begin{equation*} \begin{split} & f’ (x) = 0\\\\ & 3x^2 + 2ax + b = 0 \end{split} \end{equation*}
For x = 3
\begin{equation*} \begin{split} & 3(3)^{2} + 2a(3) + b = 0\\\\ & 27 + 6a + b = 0\\\\ & 6a + b = -27 \quad {\color {red} \dotso \: (1)} \end{split} \end{equation*}
For x = −5
\begin{equation*} \begin{split} & 3(-5)^{2} + 2a(-5) + b = 0\\\\ & 75 – 10a + b = 0\\\\ & -10a + b = -75 \quad {\color {red} \dotso \: (2)} \end{split} \end{equation*}
\begin{equation*} \begin{split} & 6a + b = -27\\\\ & -10a + b = -75 \quad (-)\\ \hline\\ & 16a = 48\\\\ & a = 3 \end{split} \end{equation*}
\begin{equation*} \begin{split} & 6a + b = – 27 \\\\ & 6(3) + b = -27 \\\\ & b = – 45 \end{split} \end{equation*}
Equation of curve
\(y = x^{3} + 3x^{2} -45x + c\)
Substitute (3, −69)
\begin{equation*} \begin{split} & y = x^{3} + 3x^{2} -45x + c\\\\ & -69 = (3)^{3} + 3(3)^{2} – 45(3) + c \\\\ & -69 = -81 + c \\\\ & c = 12 \end{split} \end{equation*}
Equation of curve
\(y = x^{3} + 3x^{2} – 45x + 12\)
Substitute (−5, m)
\begin{equation*} \begin{split} & y = x^{3} + 3x^{2} – 45x + 12\\\\ & m = (-5)^{3} + 3(-5)^{2} – 45(-5) + 12 \\\\ & m = 187 \end{split} \end{equation*}
[else]
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