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: $$x = 3, x = -2$$ Only $x=3$ lies in $[-1,4]$, so fixed point is $3$. 2. If $y = (t^2 + 2)^2$ and $t = oot{2} elax x = x^{1/2}$, find $dy/dx$. By chain rule: $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$$ Calculate each: $$y = (t^2+2)^2$$ $$\frac{dy}{dt} = 2(t^2+2)(2t) = 4t(t^2+2)$$ $$t = x^{1/2}$$ $$\frac{dt}{dx} = \frac{1}{2} x^{-1/2}$$ So: $$\frac{dy}{dx} = 4t(t^2+2) \cdot \frac{1}{2} x^{-1/2} = 2t(t^2+2) x^{-1/2}$$ Since $t = x^{1/2}$, substitute: $$\frac{dy}{dx} = 2 x^{1/2} (x + 2) x^{-1/2} = 2 (x + 2)$$ 3. Find $c$ satisfying the Mean Value Theorem for $f(x) = |x|$ on $[-2,5]$. The MVT requires $f$ to be continuous and differentiable on $(a,b)$. $f(x) = |x|$ is continuous but not differentiable at $x=0$. Calculate average rate of change: $$\frac{f(5)-f(-2)}{5 - (-2)} = \frac{5 - 2}{7} = \frac{3}{7}$$ On $(-2,0)$, $f(x) = -x$ and $f'(x) = -1$ On $(0,5)$, $f(x) = x$ and $f'(x) = 1$ Neither $-1$ nor $1$ equals $3/7$, and no $c$ in $(-2,5)$ has derivative $3/7$. So no $c$ exists in $(-2,5)$ satisfying MVT. 4. Given $f(x) = 2x^3 - 3x^2 + 4x$, find open intervals where $f$ is concave down. Find second derivative: $$f'(x) = 6x^2 - 6x + 4$$ $$f''(x) = 12x -6$$ Concave down where $f''(x) < 0$: $$12x - 6 < 0 \implies x < \frac{1}{2}$$ So $f$ is concave down on $(-\infty, \frac{1}{2})$. 5. Find intervals where $f(x) = x^3 e^x$ is increasing. First derivative via product rule: $$f'(x) = 3x^2 e^x + x^3 e^x = e^x (3x^2 + x^3) = e^x x^2 (3 + x)$$ Since $e^x > 0$ always, sign depends on $x^2$ and $(3 + x)$. $x^2 \geq 0$ and zero only at $x=0$, nonnegative everywhere. So sign depends on $(3 + x)$: Positive if: $$3 + x > 0 \implies x > -3$$ Zero at $x = -3, 0$. So $f'(x) > 0$ on $(-3,\infty)$ except at $x=0$ where derivative is zero. Therefore, $f$ is increasing on $[-3, \infty)$ but strictly increasing on $(-3, 0) \cup (0, \infty)$. Final answers: 1. Fixed point: $x=3$ 2. Derivative: $\frac{dy}{dx} = 2(x + 2)$ 3. No $c$ satisfies MVT on $[-2,5]$ 4. Concave down on $(-\infty, \frac{1}{2})$ 5. Increasing on $[-3, \infty)$