Q 30. In exercise, letg1(t)=sint,聽g2(... [FREE SOLUTION]

Chapter 12: Q 30. (page 916)

In exercise, let
$g_{1} (t) = \sin t, g_{2} (t) = \cos t, g_{3} (t) = 1 - t, f_{1} (x, y) = x^{2} + y^{2}, f_{2} (x, y) = \frac{x^{2}}{y^{2}}, f_{3} (x, y, z) = \frac{x + y}{y + z}, r_{1} (t) = 鉄� 1 + t, t - 1 鉄�, r_{2} (t) = <t, t^{2}, t^{3}>$
Either simplify the specified composition or explain why the composition cannot be formed.
$g_{1} (f_{1} (x, y), f_{2} (x, y))$

Short Answer

Expert verified

Function cannot be composed

Step by step solution

Step 1. Given information

$g_{1} (t) = \sin t, g_{2} (t) = \cos t, g_{3} (t) = 1 - t, f_{1} (x, y) = x^{2} + y^{2}, f_{2} (x, y) = \frac{x^{2}}{y^{2}}, f_{3} (x, y, z) = \frac{x + y}{y + z}, r_{1} (t) = 鉄� 1 + t, t - 1 鉄�, r_{2} (t) = <t, t^{2}, t^{3}>$

Step 2. Explanation

Domain of $g_{1}$ is R.

To compose the function $g_{1} (f_{1} (x, y), f_{2} (x, y))$ , two input variables are required.

It is against the given definition.

So, such a function cannot be composed.

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91影视

In exercise, letg1(t)=sint,g2(t)=cost,g3(t)=1-t,f1(x,y)=x2+y2,f2(x,y)=x2y2,f3(x,y,z)=x+yy+z,r1(t)=鉄�1+t,t-1鉄�,r2(t)=t,t2,t3Either simplify the specified composition or explain why the composition cannot be formed.g1f1x,y,f2x,y

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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Most popular questions from this chapter

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