Given f(x) = 5 − 4x − 3x2.

Given $f(x) = 5 - 4x - 3x^2$.

$$f(x+h) = 5 - 4(x+h) - 3(x+h)^2$$ $$f(x+h) = 5 - 4x - 4h - 3(x^2 + 2xh + h^2)$$ $$f(x+h) = 5 - 4x - 4h - 3x^2 - 6xh - 3h^2$$

$$f(x+h) - f(x) = (5 - 4x - 4h - 3x^2 - 6xh - 3h^2) - (5 - 4x - 3x^2)$$ $$f(x+h) - f(x) = 5 - 4x - 4h - 3x^2 - 6xh - 3h^2 - 5 + 4x + 3x^2$$ $$f(x+h) - f(x) = -4h - 6xh - 3h^2$$

$$\frac{f(x+h) - f(x)}{h} = \frac{-4h - 6xh - 3h^2}{h}$$ $$= \frac{h(-4 - 6x - 3h)}{h}$$ $$= -4 - 6x - 3h$$

$$\frac{dy}{dx} = \lim_{h \to 0} (-4 - 6x - 3h)$$ $$\frac{dy}{dx} = -4 - 6x - 3(0)$$ $$\frac{dy}{dx} = -4 - 6x$$

The derivative is $\boxed{-4 - 6x}$.

4. EXPONENTIAL EQUATION

If $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$, find the values of $x$, where $x \in \mathbb{R}$.

$$2^{3x+1} = 2^{3x} \cdot 2^1 = (2^x)^3 \cdot 2$$ $$2^{2x} = (2^x)^2$$ $$2^{x+1} = 2^x \cdot 2^1 = 2 \cdot 2^x$$

Substitute these into the equation:

$$2(2^x)^3 - 3(2^x)^2 + 2(2^x) = 2^x$$

Since $x \in \mathbb{R}$, $u$ must be positive ($u > 0$).
The equation becomes a cubic polynomial in $u$:

$$2u^3 - 3u^2 + 2u = u$$

$$2u^3 - 3u^2 + 2u - u = 0$$ $$2u^3 - 3u^2 + u = 0$$

Factor out $u$:

$$u(2u^2 - 3u + 1) = 0$$

This gives two possibilities:

1. $u = 0$
2. $2u^2 - 3u + 1 = 0$

For $u=0$: Since $u = 2^x$, $2^x = 0$ has no solution because $2^x$ is always positive.

For $2u^2 - 3u + 1 = 0$: Factor the quadratic equation.

$$(2u - 1)(u - 1) = 0$$

This gives two solutions for $u$:

$$2u - 1 = 0 \implies 2u = 1 \implies u = \frac{1}{2}$$ $$u - 1 = 0 \implies u = 1$$

Case 1: $u = \frac{1}{2}$

$$2^x = \frac{1}{2}$$ $$2^x = 2^{-1}$$ $$x = -1$$

Case 2: $u = 1$

$$2^x = 1$$ $$2^x = 2^0$$ $$x = 0$$

The values of $x$ are $\boxed{-1, 0}$.