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yan [13]
3 years ago
15

Find the domain and range of the function: f(t)=sec[((pi)t)/4]

Mathematics
1 answer:
beks73 [17]3 years ago
3 0

The domain and range of the function f(t)=sec[((pi)t)/4] are the following:

<span>Domain: All the real numbers except t = 2 + 4k, where k is an integer. </span>

Range: (-∞, -1] U [1, ∞)

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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Evaluate the integral by making an appropriate change of variables.
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By inspecting the integrand, the "obvious" choice for substitution would be

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<em />

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J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12

The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.

Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So

-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>

-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2

-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>

-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>

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So, the integral becomes

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=\displaystyle\frac5{14}\sin7(9^2-8^2)=\boxed{\frac{85}{14}\sin7}

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