f is acubic function defined by f(x)=ax^3+bx^2+cx+d where a.b,c and d are real numbers.Find the values at a,b,c and d ,knowing that the function has alocal maximum at point (3,3) ,local minimum at point (5,1) and apoint of inflection at point (4,2)
Cubic function
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Hello, baha jaff!
We are given three points on the cubic.
This gives us three equations:
. . \(\displaystyle \begin{array}{ccccccc}(5,1)\!: & 125a + 25b + 5c + d &=& 1 & [1] \\ (4,2)\!: & 64a + 16b + 4c + d &=& 2 & [2] \\ (3,3)\!: & 27a + 9b + 3c + d &=& 3 & [3] \end{array}\)
\(\displaystyle \text{Since (3,3) is a maximum, }f'(3) = 0\)
. . \(\displaystyle f'(x) \:=\:3ax^2 + 2bx + c \:=\:0 \quad\Rightarrow\quad f'(3) \:=\:27a + 6b + c \:=\:0\;\;[4]\;\)
Solve the system:
\(\displaystyle \begin{array}{ccccccc}\text{Subtract [1] - [2]:} & 61a + 9b + c &=& \text{-}1 & [5]\; \\ \text{Subtract [2] - [3]:} & 37a + 7b + c &=& \text{-}1 & [6]\; \\ \text{Equation [4]:} & 27a + 6b + c &=& 0 & [4]\; \end{array}\)
\(\displaystyle \begin{array}{ccccccccccc}\text{Subtract [5] - [6]:} & 24a + 2b &=& 0 & \Rightarrow & 12a + b &=& 0 & [7] \\ \text{Subtract [6] - [4]:} &&&&& 10a + b &=& \text{-}1 & [8] \end{array}\)
\(\displaystyle \text{Subtract [7] - [8]: }\;2a \:=\:1 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{2}}\)
\(\displaystyle \text{Substitute into [8]: }\:10(\tfrac{1}{2}) + b \:=\:\text{-}1 \quad\Rightarrow\quad \boxed{b \:=\:\text{-}6}\;\;\)
\(\displaystyle \text{Substitute into [4]: }\:27(\tfrac{1}{2}) + 6(\text{-}6) + c \:=\:0 \quad\Rightarrow\quad \boxed{c \,=\,\tfrac{45}{2}}\)
\(\displaystyle \text{Substitute into [3]: }\:27(\tfrac{1}{2}) + 9(\text{-}6) + 3(\tfrac{45}{2}) + d \:=\:3 \quad\Rightarrow\quad \boxed{d \,=\,\text{-}24}\)
\(\displaystyle \text{Therefore: }\;f(x) \;=\;\tfrac{1}{2}x^3 - 6x^2 + \tfrac{45}{2}x - 24\)
We are given three points on the cubic.
This gives us three equations:
. . \(\displaystyle \begin{array}{ccccccc}(5,1)\!: & 125a + 25b + 5c + d &=& 1 & [1] \\ (4,2)\!: & 64a + 16b + 4c + d &=& 2 & [2] \\ (3,3)\!: & 27a + 9b + 3c + d &=& 3 & [3] \end{array}\)
\(\displaystyle \text{Since (3,3) is a maximum, }f'(3) = 0\)
. . \(\displaystyle f'(x) \:=\:3ax^2 + 2bx + c \:=\:0 \quad\Rightarrow\quad f'(3) \:=\:27a + 6b + c \:=\:0\;\;[4]\;\)
Solve the system:
\(\displaystyle \begin{array}{ccccccc}\text{Subtract [1] - [2]:} & 61a + 9b + c &=& \text{-}1 & [5]\; \\ \text{Subtract [2] - [3]:} & 37a + 7b + c &=& \text{-}1 & [6]\; \\ \text{Equation [4]:} & 27a + 6b + c &=& 0 & [4]\; \end{array}\)
\(\displaystyle \begin{array}{ccccccccccc}\text{Subtract [5] - [6]:} & 24a + 2b &=& 0 & \Rightarrow & 12a + b &=& 0 & [7] \\ \text{Subtract [6] - [4]:} &&&&& 10a + b &=& \text{-}1 & [8] \end{array}\)
\(\displaystyle \text{Subtract [7] - [8]: }\;2a \:=\:1 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{2}}\)
\(\displaystyle \text{Substitute into [8]: }\:10(\tfrac{1}{2}) + b \:=\:\text{-}1 \quad\Rightarrow\quad \boxed{b \:=\:\text{-}6}\;\;\)
\(\displaystyle \text{Substitute into [4]: }\:27(\tfrac{1}{2}) + 6(\text{-}6) + c \:=\:0 \quad\Rightarrow\quad \boxed{c \,=\,\tfrac{45}{2}}\)
\(\displaystyle \text{Substitute into [3]: }\:27(\tfrac{1}{2}) + 9(\text{-}6) + 3(\tfrac{45}{2}) + d \:=\:3 \quad\Rightarrow\quad \boxed{d \,=\,\text{-}24}\)
\(\displaystyle \text{Therefore: }\;f(x) \;=\;\tfrac{1}{2}x^3 - 6x^2 + \tfrac{45}{2}x - 24\)
