91 lines
2.2 KiB
TeX
91 lines
2.2 KiB
TeX
\beginsection 17.3
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Accept on faith that the following familiar functions are continuous on
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their domains: $\sin x$, $\cos x$, $e^x$, $2^x$, $\log_e x$ for
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$x>0$, $x^p$ for $x>0$ [$p$ any real number].
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Use these facts and theorems in this section to prove that the following
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functions are continuous.
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\medskip
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\noindent
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(a) $\log_e(1+\cos^4 x)$
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\medskip
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\itemitem{}
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The function $\cos^4 x$ is continuous by Theorem 17.4 (ii),
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product of continuous functions.
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The function $1+\cos^4 x$ is continuous by Theorem 17.4 (i),
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sum of continuous functions.
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The function $\log_e(1+\cos^4 x)$ is continuous by Theorem 17.5,
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composition of continuous functions.
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\medskip
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\noindent
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(b) $[\sin^2x+\cos^6x]^\pi$
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\medskip
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\itemitem{}
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This is equivalent to $\exp[\pi\log_e(\sin^2x+\cos^6x)]$ which is continuous by
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Theorems 17.4 and 17.5.
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Now show that $\sin^2x+\cos^6x>0$ to ensure the domain requirement of $\log_e$.
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We have $\sin^2x\ge0$, $\cos^6x\ge0$ by even exponents.
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By $\sin^2x+\cos^2x=1$, $\sin^2x$ and $\cos^2x$ cannot both be zero for the
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same $x$.
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If $\cos^2x\ne0$ then $\cos^6x\ne0$.
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Therefore $\sin^2x+\cos^6x>0$.
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\medskip
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\noindent
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(c) $2^{x^2}$
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\medskip
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\itemitem{}
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The function $x^2$ is continuous by $x^2=xx$ and
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Theorem 17.4 (ii), product of continuous functions.
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By Theorem 17.5, composition of continuous fuctions, $2^{x^2}$ is continuous.
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\medskip
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\noindent
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(d) $8^x$
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\medskip
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\itemitem{}
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This is equivalent to $\exp(x\log_e8)$ which is continuous by Theorems 17.4 and 17.5.
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\medskip
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\noindent
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(e) $\tan x$ for $x\ne$ odd multiple of $\pi/2$.
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\medskip
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\itemitem{}
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$\tan x=\sin x/\cos x$ is continuous by Theorem 17.4 (iii), ratio of
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continuous functions.
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The odd multiple restriction ensures $\cos x\ne0$.
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\medskip
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\noindent
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(f) $x\sin(1/x)$ for $x\ne0$
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\medskip
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\itemitem{}
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$1/x=\exp(-\log_ex)$ is continuous for $x>0$.
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For $x<0$, $1/x=-1/|x|$ so $1/x$ is also continous for $x<0$.
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Finally, $x\sin(1/x)$ is continuous by Theorems 17.4 and 17.5.
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\medskip
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\noindent
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(g) $x^2\sin(1/x)$ for $x\ne0$
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\medskip
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\itemitem{}
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$x^2\sin(1/x)=x[x\sin(1/x)]$, see (f).
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\medskip
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\noindent
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(h) $(1/x)\sin(1/x^2)$ for $x\ne0$
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\medskip
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\itemitem{}
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The only thing new here is $1/x^2$ which is continuous by $1/x^2=(1/x)(1/x)$
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and Theorem 17.4 (ii). See (f) for continuity of $1/x$.
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