Problem 60 Determine whether \(h=g \circ f\... [FREE SOLUTION]

Chapter 0: Problem 60

Determine whether \(h=g \circ f\) is even, odd, or neither, given that a. both \(g\) and \(f\) are even. b. \(g\) is even and \(f\) is odd. c. \(g\) is odd and \(f\) is even. d. both \(g\) and \(f\) are odd.

Short Answer

Expert verified

a. \(h\) is even. b. \(h\) is even. c. \(h\) is odd. d. \(h\) is odd.

Step by step solution

a. Both \(g\) and \(f\) are even.

Since both \(g\) and \(f\) are even functions, by definition we have: 1. \(g(-x)=g(x)\) 2. \(f(-x)=f(x)\) Now let's find \(h(-x)\): \(h(-x) = g(f(-x)) = g(f(x)) = h(x)\) Since \(h(-x) = h(x)\), we can conclude that \(h\) is even.

b. \(g\) is even and \(f\) is odd.

Now, since \(g\) is even and \(f\) is odd functions, by definition we have: 1. \(g(-x)=g(x)\) 2. \(f(-x)=-f(x)\) Now let's find \(h(-x)\): \(h(-x) = g(f(-x)) = g(-f(x))\) Because \(g\) is even, we get: \(h(-x) = g(f(x)) = h(x)\) Since \(h(-x) = h(x)\), we can conclude that \(h\) is even.

c. \(q\) is odd and \(f\) is even (assuming \(q\) should be \(g\)).

Now, since \(g\) is odd and \(f\) is even functions, by definition we have: 1. \(g(-x)=-g(x)\) 2. \(f(-x)=f(x)\) Now let's find \(h(-x)\): \(h(-x) = g(f(-x)) = g(f(x))\) Because \(g\) is odd, we get: \(h(-x) = -g(f(x)) = -h(x)\) Since \(h(-x) = -h(x)\), we can conclude that \(h\) is odd.

d. Both \(g\) and \(f\) are odd.

Since both \(g\) and \(f\) are odd functions, by definition we have: 1. \(g(-x)=-g(x)\) 2. \(f(-x)=-f(x)\) Now let's find \(h(-x)\): \(h(-x) = g(f(-x)) = g(-f(x))\) Because both \(g\) and \(f\) are odd, we get: \(h(-x) = -g(f(x)) = -h(x)\) Since \(h(-x) = -h(x)\), we can conclude that \(h\) is odd.

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91影视

Determine whether \(h=g \circ f\) is even, odd, or neither, given that a. both \(g\) and \(f\) are even. b. \(g\) is even and \(f\) is odd. c. \(g\) is odd and \(f\) is even. d. both \(g\) and \(f\) are odd.

Short Answer

Step by step solution

a. Both \(g\) and \(f\) are even.

b. \(g\) is even and \(f\) is odd.

c. \(q\) is odd and \(f\) is even (assuming \(q\) should be \(g\)).

d. Both \(g\) and \(f\) are odd.

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