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SVETLANKA909090 [29]
3 years ago
10

Help!!!!! Suppose f(x) = x2. What is the graph of g(x) = f(3x)?

Mathematics
1 answer:
STALIN [3.7K]3 years ago
7 0

Answer:

we conclude that g(x) = 9x² is the result of f(x) vertically stretched by multiplying each of its y-coordinates by 9.

Also, check the attached graph is matched with option B.

Hence, option (B) is the correct graph.

Step-by-step explanation:

Given

  • f(x) = x²
  • g(x) = f(3x)

so

  • g(x) = (3x)² = 9x²

Thus, the graph of g(x) = 9x² is shown below.

From the graph, it is clear that at x = 0, y = 0

So, the y-intercept is at (0, 0).

Clearly

  • f(x) = x²
  • g(x) = 9x²

Note that the vertex of the graph also lies at (0, 0), because it is clear that g(x) = 9x² is the result of f(x) vertically stretched by multiplying each of its y-coordinates by 9.

Since 9 > 0, it means g(x) = 9x² is vertically stretched by multiplying each of its y-coordinates by 9.

Therefore, we conclude that g(x) = 9x² is the result of f(x) vertically stretched by multiplying each of its y-coordinates by 9.

Also, check the attached graph is matched with option B.

Hence, option (B) is the correct graph.

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Ann [662]

Answer:

Required series is:

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

Step-by-step explanation:

Given that

                           f'(x) = -\frac{1}{1 + x^{2}} ---(1)

We know that:

                  \frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}} ---(2)

Comparing (1) and (2)

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Using power series expansion for tan^{-1}x

f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx

= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]

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as

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Hence,

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

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