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2 years ago

Let f(x)=6/5x−1/2. Describe the transformations from the graph of f to the graphs of g(x)=−f(x) and h(x)=f(−x).

Mathematics

1 answer:

aivan3 [116]2 years ago

8 0

Answer:

g(x)=-6/5x+1/2

h(x)=-6/5x-1/2

Step-by-step explanation:

1). g(x)=−f(x) ?

f(x)=6/5x−1/2

g(x)=−(6/5x−1/2)

2). h(x)=f(−x) ?

f(-x)=6/5(-x)−1/2

f(-x)=-6/5x-1/2

h(x)=-6/5x-1/2

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Working based on the information given, the mean values of each group with with men having an average score of 1.02 and women have an average of 1.79 this reveals that distressing nausea on average is higher in women than in men . However, to test if there is a significant difference would be challenging as the information given isn't enough to make proceed with the test as the standard deviations of the two groups aren't given and no accompanying sample data is given.

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3 years ago

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34 classes

Step-by-step explanation:

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3 years ago

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Answer:

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3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

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(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

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c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

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3 years ago

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Gre4nikov [31]

Answer:
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Step by step explanation:
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3 years ago