This problem can be solved using the Intermediate Value Theorem, which states that if a continuous function $f: [a,b] \rightarrow \mathbb{R}$ takes on values $f(a)$ and $f(b)$ at the endpoints of an interval $[a,b]$, then it takes on every value between $f(a)$ and $f(b)$ at least once in the interval.

Since $f$ is continuous and $f(a) \cdot f(b) < 0$, we can assume without loss of generality that $f(a) > 0$ and $f(b) < 0$. This is because if $f(a) < 0$ and $f(b) > 0$, we can simply consider the function $-f$, which is also continuous and satisfies $(-f)(a) \cdot (-f)(b) < 0$.

Now, consider the function $g(x) = f(x) – k$, where $k$ is a constant chosen such that $g(a) = f(a) – k > 0$ and $g(b) = f(b) – k < 0$. Since $f$ is continuous, $g$ is also continuous. By the Intermediate Value Theorem, there exists a point $c \in (a,b)$ such that $g(c) = 0$, which means $f(c) – k = 0$, or equivalently, $f(c) = k$.

Since $k$ was chosen such that $g(a) = f(a) – k > 0$ and $g(b) = f(b) – k < 0$, we have $f(a) > k$ and $f(b) < k$. Therefore, $f(a) > f(c) > f(b)$, which means that $f$ takes on every value between $f(a)$ and $f(b)$ at least once in the interval $(a,b)$. In particular, $f(c) = 0$ for some $c \in (a,b)$, as desired.

To use the bisection method, we first need to find an interval [a,b] that contains a root of the equation. We can do this by evaluating the function at some points and looking for a sign change. Let’s try evaluating the function at x=0, x=1, and x=2:

f(0) = 0^5 – 3(0)^4 + 2(0)^3 – 0^2 + 0 = 0 f(1) = 1^5 – 3(1)^4 + 2(1)^3 – 1^2 + 1 = 0 f(2) = 2^5 – 3(2)^4 + 2(2)^3 – 2^2 + 2 = 18

We see that f(0) and f(1) have opposite signs, so there must be a root between 0 and 1. Let’s use the bisection method to find this root:

First iteration: a = 0, b = 1, c = (a+b)/2 = 0.5, f(c) = -0.015625 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Second iteration: a = 0, b = 0.5, c = (a+b)/2 = 0.25, f(c) = -0.1826171875 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Third iteration: a = 0, b = 0.25, c = (a+b)/2 = 0.125, f(c) = 0.2136229277 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Fourth iteration: a = 0.125, b = 0.25, c = (a+b)/2 = 0.1875, f(c) = 0.0282497385 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Fifth iteration: a = 0.125, b = 0.1875, c = (a+b)/2 = 0.15625, f(c) = -0.0925126744 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Sixth iteration: a = 0.125, b = 0.15625, c = (a+b)/2 = 0.140625, f(c) = -0.0314813204 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Seventh iteration: a = 0.125, b = 0.140625, c = (a+b)/2 = 0.1328125, f(c) = -0.0012603379 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Eighth iteration: a = 0.125, b = 0.1328125, c = (a+b)/2 = 0.12890625, f(c) = 0.0139563582 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Ninth iteration: a = 0.12890625, b = 0.1328125, c = (a+b)/2

First equation: $x=\cos x$

Bisection Method: Let $f(x)=x-\cos x$. Then, $f$ is continuous and changes sign on the interval $[0,1]$ since $f(0)=1>0$ and $f(1)=\cos 1-1<0$. Starting with the interval $[a_0,b_0]=[0,1]$, we have:

• $c_0=\frac{a_0+b_0}{2}=\frac{1}{2}$, $f(c_0)=c_0-\cos c_0\approx 0.0707>0$,
• $a_1=a_0$, $b_1=c_0$ since $f(a_0)f(c_0)<0$,
• $c_1=\frac{a_1+b_1}{2}=\frac{1}{4}$, $f(c_1)\approx -0.3065<0$,
• $a_2=c_1$, $b_2=b_1$ since $f(c_1)f(b_1)<0$,
• $c_2=\frac{a_2+b_2}{2}=\frac{3}{8}$, $f(c_2)\approx -0.1207<0$,
• $a_3=c_2$, $b_3=b_2$ since $f(c_2)f(b_2)<0$,
• $c_3=\frac{a_3+b_3}{2}=\frac{7}{16}$, $f(c_3)\approx -0.0247<0$,
• $a_4=c_3$, $b_4=b_3$ since $f(c_3)f(b_3)<0$,
• $c_4=\frac{a_4+b_4}{2}=\frac{15}{32}$, $f(c_4)\approx 0.0230>0$,
• $a_5=a_4$, $b_5=c_4$ since $f(a_4)f(c_4)<0$,
• $c_5=\frac{a_5+b_5}{2}=\frac{23}{64}$, $f(c_5)\approx -0.0016<0$,
• $a_6=c_5$, $b_6=b_5$ since $f(c_5)f(b_5)<0$,
• $c_6=\frac{a_6+b_6}{2}=\frac{87}{256}$, $f(c_6)\approx 0.0105>0$,
• $a_7=a_6$, $b_7=c_6$ since $f(a_6)f(c_6)<0$,
• $c_7=\frac{a_7+b_7}{2}=\frac{169}{512}$, $f(c_7)\approx 0.0044>0$,
• $a_8=a_7$, $b_8=c_7$ since $f(a_7)f(c_7)<0$,
• $c_8=\frac{a_8+b_8}{2}=\frac{325}{1024}$, $f(c_8)\approx 0.0014>0$,
• $a_9=a_8$, $b_9=c_8$ since $f(a_8)f(c_8)<0$,
• $c_9=\frac{a_9+b_9}{2}=\frac