1. **Find the fixed points for the function $f(x) = x^2 - 6$ in the interval $[-1,4]$**. A fixed point is where $f(x) = x$. So, solve: $$x = x^2 - 6$$ Rearrange: $$x^2 - x - 6 = 0$$ Factor: $$ (x - 3)(x + 2) = 0 $$ Solutions: $$x = 3 \quad \text{or} \quad x = -2$$ Check interval $[-1,4]$: $x=3$ is in the interval, $x=-2$ is not. 2. **If $y = (t^2 + 2)^2$ and $t = x^{1/3}$, find $\frac{dy}{dx}$**. Use chain rule: $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$$ Calculate: $$\frac{dy}{dt} = 2(t^2 + 2)(2t) = 4t(t^2 + 2)$$ $$\frac{dt}{dx} = \frac{d}{dx} x^{1/3} = \frac{1}{3} x^{-2/3}$$ So: $$\frac{dy}{dx} = 4t(t^2 + 2) \cdot \frac{1}{3} x^{-2/3} = \frac{4}{3} t (t^2 + 2) x^{-2/3}$$ Recall $t = x^{1/3}$, substitute: $$\frac{dy}{dx} = \frac{4}{3} x^{1/3} (x^{2/3} + 2) x^{-2/3} = \frac{4}{3} x^{1/3 - 2/3} (x^{2/3} + 2) = \frac{4}{3} x^{-1/3} (x^{2/3} + 2)$$ 3. **Find $c$ satisfying the Mean Value Theorem (MVT) for $f(x) = |x|$ on $[-2,5]$**. Calculate average rate of change: $$\frac{f(5) - f(-2)}{5 - (-2)} = \frac{5 - 2}{7} = \frac{3}{7}$$ Note $f(x) = |x|$ is not differentiable at $x=0$, derivative left side $-1$, right side $1$. Derivative: $$f'(x) = \begin{cases} -1 & x < 0 \\ 1 & x > 0 \end{cases}$$ No $c$ in $(-2,5)$ gives $f'(c) = \frac{3}{7}$ between $-1$ and $1$, so no $c$ satisfies MVT. 4. **Find open interval(s) where $f(x) = 2x^3 - 3x^2 + 4x$ is concave down**. Find second derivative: $$f'(x) = 6x^2 - 6x + 4$$ $$f''(x) = 12x - 6$$ Concave down means: $$f''(x) < 0 \Rightarrow 12x - 6 < 0 \Rightarrow x < \frac{1}{2}$$ So $f$ is concave down on interval $(-\infty, \frac{1}{2})$. 5. **Find interval(s) where $f(x) = x^3 e^x$ is increasing**. Find derivative using product rule: $$f'(x) = \frac{d}{dx}(x^3) e^x + x^3 \frac{d}{dx}(e^x) = 3x^2 e^x + x^3 e^x = e^x (3x^2 + x^3) = e^x x^2 (3 + x)$$ Since $e^x > 0$ for all $x$, sign depends on $x^2 (3 + x)$. $x^2 \geq 0$ always, zero at $x=0$. Term $3+x$ > 0 for $x > -3$, < 0 for $x < -3$. Hence, $f'(x) > 0$ when $x^2 (3+x) > 0$, i.e., when $x eq 0$ and $x > -3$. At $x=0$, $f'(0) = 0$. So the function is increasing on $(-3,0) \cup (0, \infty)$. **Final Answers:** 1. Fixed point in $[-1,4]$ is $x = 3$. 2. $$\frac{dy}{dx} = \frac{4}{3} x^{-1/3} (x^{2/3} + 2)$$ 3. No $c$ satisfies MVT for $f(x) = |x|$ on $[-2,5]$. 4. Concave down on $(-\infty, \frac{1}{2})$. 5. Increasing on $(-3,0) \cup (0, \infty)$.