lugreget9
2021-12-17
Answered

How do you find $f\left(g(-3)\right)$ given $f\left(x\right)=2x-1$ and $g\left(x\right)=3x$ and $h\left(x\right)={x}^{2}+1$

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braodagxj

Answered 2021-12-18
Author has **38** answers

First, find g(−3) by substituting −3 for each occurrence of x in g(x):

g(x)=3x becomes:

g(-3)=3 * -3

g(-3)=-9

Therefore: f(g(-3))=f(-9)

Because g(-3)=-9 then

To find f(-9) we can substitute -9 we can substitute x in f(x)

f(x)=2x-1 becomes

f(-9)=(2 * -9)-1

f(-9)=-18-1

f(-9)=-19

f(g(-3))=-19

g(x)=3x becomes:

g(-3)=3 * -3

g(-3)=-9

Therefore: f(g(-3))=f(-9)

Because g(-3)=-9 then

To find f(-9) we can substitute -9 we can substitute x in f(x)

f(x)=2x-1 becomes

f(-9)=(2 * -9)-1

f(-9)=-18-1

f(-9)=-19

f(g(-3))=-19

Mason Hall

Answered 2021-12-19
Author has **36** answers

Another solution?

asked 2021-09-06

I must have made a mistake in finding the composite functions $f\circ g{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}g\circ f,$ because I notice that $f\circ g$ is not the same function as $g\circ f$ .Determine whether the statement makes sense or does not make sense, and explain your reasoning.

asked 2022-09-14

Thorem: If $f(x)$ is continuous at $L$ and $\underset{x\to a}{lim}g(x)=L$, then $\underset{x\to a}{lim}f(g(x)=f(\underset{x\to a}{lim}g(x))=f(L)$.

Proof: Assume $f(x)$ is continuous at a point $L$, and that $\underset{x\to a}{lim}g(x)=L$.

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}\delta >0:[|x-L|<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}f(x)-f(L){\textstyle |}<\epsilon ]$.

And $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|g(x)-L<\delta ]$.

So, $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(g(x))-f(L)|<\u03f5]$.

$\underset{x\to a}{lim}g(x)=L$ so $f(\underset{x\to a}{lim}g(x))=f(L)$.

Proof: Assume $f(x)$ is continuous at a point $L$, and that $\underset{x\to a}{lim}g(x)=L$.

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}\delta >0:[|x-L|<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}f(x)-f(L){\textstyle |}<\epsilon ]$.

And $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|g(x)-L<\delta ]$.

So, $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(g(x))-f(L)|<\u03f5]$.

$\underset{x\to a}{lim}g(x)=L$ so $f(\underset{x\to a}{lim}g(x))=f(L)$.

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