Question

for each of problems 1-5, determine whether or not the intermediate value theorem can be applied to the function on the given interval. if it applies, explain how we know there exists a zero of the function on the interval. if it does not apply, explain why. no calculators!!

1. $f(x)=x^3 + x - 1$ on $0,1$
2. $f(x)=\

$$\begin{cases}x - 5, & x\\leq2 \\\\ x^2 - 1, & x > 2\\end{cases}$$

$ on $1,3$

1. $f(x)=x^2 + x - 1$ on $0,5$
2. $f(x)=x^2 - x - \cos x$ on $0,\pi$
3. $f(x)=\

$$\begin{cases}2 - x, & x\\leq0 \\\\ 2 - x^2, & x > 0\\end{cases}$$

$ on $-1,2$

1. the functions $f$ and $g$ are continuous for all real numbers. the table below gives values of the functions at selected values of $x$. the function $h(x)=f(g(x)) - 6$.

\

$$\begin{tabular}{|c|c|c|} \\hline $x$ & $f(x)$ & $g(x)$ \\\\ \\hline 1 & 6 & 2 \\\\ \\hline 2 & 9 & 3 \\\\ \\hline 3 & 10 & 4 \\\\ \\hline 4 & -1 & 6 \\\\ \\hline \\end{tabular}$$

Explanation:

Problem 1

Step1: Check continuity on [0,1]

Polynomials are continuous everywhere, so $f(x)=x^3+x-1$ is continuous on $[0,1]$.

Step2: Compute f(0) and f(1)

$f(0)=0^3+0-1=-1$
$f(1)=1^3+1-1=1$

Step3: Apply IVT criteria

Since $f(0)=-1<0$ and $f(1)=1>0$, by IVT, there exists a $c\in(0,1)$ where $f(c)=0$.

Step1: Check continuity at x=2

Left-hand limit: $\lim_{x\to2^-}f(x)=2-5=-3$
Right-hand limit: $\lim_{x\to2^+}f(x)=2^2-1=3$

Step2: Verify continuity failure

$\lim_{x\to2^-}f(x)
eq\lim_{x\to2^+}f(x)$, so $f(x)$ is discontinuous at $x=2\in[1,3]$.

Step1: Check continuity on [0,5]

Polynomials are continuous everywhere, so $f(x)=x^2+x-1$ is continuous on $[0,5]$.

Step2: Compute f(0) and f(5)

$f(0)=0^2+0-1=-1$
$f(5)=5^2+5-1=29$

Step3: Apply IVT criteria

Since $f(0)=-1<0$ and $f(5)=29>0$, by IVT, there exists a $c\in(0,5)$ where $f(c)=0$.

Answer:

The Intermediate Value Theorem applies. $f(x)$ is continuous on $[0,1]$, $f(0)=-1<0$, and $f(1)=1>0$, so there is a zero in $(0,1)$.

---

Problem 2