Sure, here are the solutions:

a) (f+g)(x) = f(x) + g(x) = (3x - 4) + (2x² + 9) = 2x² + 3x + 5

b) (f-g)(x) = f(x) - g(x) = (3x - 4) - (2x² + 9) = -2x² + 3x - 13

c) g(x) = 2x² + 9

d) (f=g)(x) is not a valid mathematical operation. If you meant f(x) = g(x), then you are setting the two functions equal to each other and solving for x.

e) (g-1)(x) is not a valid mathematical operation. If you meant g(x) - 1, then it would be (2x² + 9) - 1 = 2x² + 8.

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Sure, here are the solutions:

a) (f+g)(x) = f(x) + g(x) = (3x - 4) + (2x² + 9) = 2x² + 3x + 5

b) (f-g)(x) = f(x) - g(x) = (3x - 4) - (2x² + 9) = -2x² + 3x - 13

c) g(x) = 2x² + 9

d) (f=g)(x) is not a valid mathematical operation. If you meant f(x) = g(x), then you are setting the two functions equal to each other and solving for x.

e) (g-1)(x) is not a valid mathematical operation. If you meant g(x) - 1, then it would be (2x² + 9) - 1 = 2x² + 8.

Knowee AI · Accepted Answer

Sure, here are the solutions:

a) (f+g)(x) = f(x) + g(x) = (3x - 4) + (2x² + 9) = 2x² + 3x + 5

b) (f-g)(x) = f(x) - g(x) = (3x - 4) - (2x² + 9) = -2x² + 3x - 13

c) g(x) = 2x² + 9

d) (f=g)(x) is not a valid mathematical operation. If you meant f(x) = g(x), then you are setting the two functions equal to each other and solving for x.

e) (g-1)(x) is not a valid mathematical operation. If you meant g(x) - 1, then it would be (2x² + 9) - 1 = 2x² + 8.

If f(x)=3x-4, g(x)=2x²+9, then finda) (f+g)(x) b) (f-g)(x)c) (g(x)d) (f=g)(x)e) (g-1)(x)

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