Consider a cubic, $f(x) = ax^3 + bx^2 + cx + 4$, $a, b, c \in R$ and $f''\left(\dfrac{-2}{3}\right) = 0$, $f'(0) = 3$, $f\left(\dfrac{-2}{3}\right) = \dfrac{5}{3}$, and $g = f^{-1}$, then:
$a + b = c$
$abc = 6$
$(g(x) \cdot f(g(x)))' \big|_{x=4} = \dfrac{4}{3}$
$(g(x) \cdot g(f(x)))' \big|_{x=4} = \dfrac{3}{4}$
Step-by-Step Solution
Key Concept: The core idea is to first simplify expressions involving composite functions of an inverse function, i.e., $f(g(x)) = x$ and $g(f(x)) = x$, before applying differentiation rules. Additionally, effectively using the derivative of an inverse function, $g'(x) = 1/f'(g(x))$, and identifying specific function values like $f(0)=4 \Rightarrow g(4)=0$ are crucial.
Step 1: Determine the coefficients $a, b, c$.
Given the cubic function $f(x) = ax^3 + bx^2 + cx + 4$.
The first derivative is $f'(x) = 3ax^2 + 2bx + c$.
The second derivative is $f''(x) = 6ax + 2b$.
Using the condition $f''\left(\dfrac{-2}{3}\right) = 0$:
$$6a\left(\dfrac{-2}{3}\right) + 2b = 0$$
$$-4a + 2b = 0$$
$$b = 2a$$
Using the condition $f'(0) = 3$:
$$3a(0)^2 + 2b(0) + c = 3$$
$$c = 3$$
Using the condition $f\left(\dfrac{-2}{3}\right) = \dfrac{5}{3}$:
Substitute $b=2a$ and $c=3$ into the function $f(x)$:
$$a\left(\dfrac{-2}{3}\right)^3 + b\left(\dfrac{-2}{3}\right)^2 + c\left(\dfrac{-2}{3}\right) + 4 = \dfrac{5}{3}$$
$$a\left(\dfrac{-8}{27}\right) + (2a)\left(\dfrac{4}{9}\right) + (3)\left(\dfrac{-2}{3}\right) + 4 = \dfrac{5}{3}$$
$$\dfrac{-8a}{27} + \dfrac{8a}{9} - 2 + 4 = \dfrac{5}{3}$$
To combine the terms with $a$, find a common denominator:
$$\dfrac{-8a}{27} + \dfrac{24a}{27} + 2 = \dfrac{5}{3}$$
$$\dfrac{16a}{27} = \dfrac{5}{3} - 2$$
$$\dfrac{16a}{27} = \dfrac{5}{3} - \dfrac{6}{3}$$
$$\dfrac{16a}{27} = \dfrac{-1}{3}$$
$$a = \dfrac{-1}{3} \cdot \dfrac{27}{16}$$
$$a = \dfrac{-9}{16}$$
Now, determine $b$ using $b=2a$:
$$b = 2\left(\dfrac{-9}{16}\right) = \dfrac{-9}{8}$$
Thus, the coefficients of the cubic function are $a = \dfrac{-9}{16}$, $b = \dfrac{-9}{8}$, and $c = 3$.
Step 2: Evaluate the given propositions.
**Proposition 1: $a + b = c$**
Substitute the determined values of $a, b, c$:
$$a + b = \dfrac{-9}{16} + \dfrac{-9}{8} = \dfrac{-9}{16} - \dfrac{18}{16} = \dfrac{-27}{16}$$
Since $c = 3$, and $\dfrac{-27}{16} \neq 3$, this proposition is false.
**Proposition 2: $abc = 6$**
Substitute the determined values of $a, b, c$:
$$abc = \left(\dfrac{-9}{16}\right)\left(\dfrac{-9}{8}\right)(3) = \left(\dfrac{81}{128}\right)(3) = \dfrac{243}{128}$$
Since $\dfrac{243}{128} \neq 6$, this proposition is false.
**Proposition 3: $(g(x) \cdot f(g(x)))' \big|_{x=4} = \dfrac{4}{3}$**
Let $y(x) = g(x) \cdot f(g(x))$.
Since $g = f^{-1}$, by definition $f(g(x)) = x$.
Therefore, $y(x) = g(x) \cdot x$.
Differentiating $y(x)$ with respect to $x$ using the product rule:
$$y'(x) = g'(x) \cdot x + g(x) \cdot 1$$
To evaluate $y'(4)$, we need the values of $g(4)$ and $g'(4)$.
To find $g(4)$: Since $g = f^{-1}$, $g(4) = k$ implies $f(k) = 4$.
From the definition of $f(x)$, $f(0) = a(0)^3 + b(0)^2 + c(0) + 4 = 4$.
Thus, $f(0) = 4$, which implies $g(4) = 0$.
To find $g'(4)$: Using the inverse function theorem, $g'(x) = \dfrac{1}{f'(g(x))}$.
So, $g'(4) = \dfrac{1}{f'(g(4))} = \dfrac{1}{f'(0)}$.
We know $f'(x) = 3ax^2 + 2bx + c$.
Evaluating $f'(0)$: $f'(0) = 3a(0)^2 + 2b(0) + c = c$.
From Step 1, $c = 3$. So, $f'(0) = 3$.
Therefore, $g'(4) = \dfrac{1}{3}$.
Substitute these values into the expression for $y'(4)$:
$$y'(4) = g'(4) \cdot 4 + g(4) = \left(\dfrac{1}{3}\right)(4) + 0 = \dfrac{4}{3}$$
This proposition is true.
**Proposition 4: $(g(x) \cdot g(f(x)))' \big|_{x=4} = \dfrac{3}{4}$**
Let $z(x) = g(x) \cdot g(f(x))$.
Since $g = f^{-1}$, by definition $g(f(x)) = x$.
Therefore, $z(x) = g(x) \cdot x$.
This is the same expression as $y(x)$ in Proposition 3.
Differentiating $z(x)$ with respect to $x$:
$$z'(x) = g'(x) \cdot x + g(x)$$
Evaluating at $x=4$:
$$z'(4) = g'(4) \cdot 4 + g(4) = \left(\dfrac{1}{3}\right)(4) + 0 = \dfrac{4}{3}$$
Since $\dfrac{4}{3} \neq \dfrac{3}{4}$, this proposition is false.
Correct Answer: 1, 2, 3