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Let f be continuous over a closed, bounded interval $\left[a,b\right].$ If z is any real number between $f\left(a\right)$ and $f\left(b\right),$ then there is a number c in $\left[a,b\right]$ satisfying $f\left(c\right)=z$ in [link] .
Show that $f\left(x\right)=x-\text{cos}\phantom{\rule{0.1em}{0ex}}x$ has at least one zero.
Since $f\left(x\right)=x-\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous over $\left(\text{\u2212}\infty ,\text{+}\infty \right),$ it is continuous over any closed interval of the form $\left[a,b\right].$ If you can find an interval $\left[a,b\right]$ such that $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in $\left(a,b\right)$ that satisfies $f\left(c\right)=0.$ Note that
and
Using the Intermediate Value Theorem, we can see that there must be a real number c in $\left[0,\pi \text{/}2\right]$ that satisfies $f\left(c\right)=0.$ Therefore, $f\left(x\right)=x-\text{cos}\phantom{\rule{0.1em}{0ex}}x$ has at least one zero.
If $f\left(x\right)$ is continuous over $\left[0,2\right],f\left(0\right)>0$ and $f\left(2\right)>0,$ can we use the Intermediate Value Theorem to conclude that $f\left(x\right)$ has no zeros in the interval $[0,2\text{]?}$ Explain.
No. The Intermediate Value Theorem only allows us to conclude that we can find a value between $f\left(0\right)$ and $f\left(2\right);$ it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function $f\left(x\right)={\left(x-1\right)}^{2}.$ It satisfies $f\left(0\right)=1>0,f\left(2\right)=1>0,$ and $f\left(1\right)=0.$
For $f\left(x\right)=1\text{/}x,f\left(\mathrm{-1}\right)=\mathrm{-1}<0$ and $f\left(1\right)=1>0.$ Can we conclude that $f\left(x\right)$ has a zero in the interval $\left[\mathrm{-1},1\right]?$
No. The function is not continuous over $\left[\mathrm{-1},1\right].$ The Intermediate Value Theorem does not apply here.
Show that $f\left(x\right)={x}^{3}-{x}^{2}-3x+1$ has a zero over the interval $\left[0,1\right].$
$f\left(0\right)=1>0,f\left(1\right)=\mathrm{-2}<0;f\left(x\right)$ is continuous over $\left[0,1\right].$ It must have a zero on this interval.
For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.
$f\left(x\right)=\frac{1}{\sqrt{x}}$
The function is defined for all x in the interval $\left(0,\infty \right).$
$f\left(x\right)=\frac{2}{{x}^{2}+1}$
$f\left(x\right)=\frac{x}{{x}^{2}-x}$
Removable discontinuity at $x=0;$ infinite discontinuity at $x=1$
$g\left(t\right)={t}^{\mathrm{-1}}+1$
$f\left(x\right)=\frac{5}{{e}^{x}-2}$
Infinite discontinuity at $x=\text{ln}\phantom{\rule{0.1em}{0ex}}2$
$f\left(x\right)=\frac{\left|x-2\right|}{x-2}$
$H\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}2x$
Infinite discontinuities at $x=\frac{\left(2k+1\right)\pi}{4},$ for $k=0,\pm 1,\pm 2,\pm 3\text{,\u2026}$
$f\left(t\right)=\frac{t+3}{{t}^{2}+5t+6}$
For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?
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