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>>604312907 Where the fuck are you doing this homework? You're not in an english speaking country are you? I've only heard of bilingual schools in latin american countries, and this material is basic so you're doing middle school grade shit.
>>604314960 No it's not, you faggot. The image of a function evaluated at a point is the corresponding member of the function's co-domain.
>>604314940 This guy's got it. To find the image, simply plug in the number it gives you for x.
For 3, simply plug in (x+h) for x. You'll get (x+h)^3+(x+h), then use binomial theorem (or wolframalpha) to expand.
For the fourth, evaluate the function at x = -1 for a, then plug in a+b for b), then plug in x-h for c).
For number five, it's asking you to solve 5 = x^2+x-1, which simplifies to 0 = x^2 + x - 6, which factors to (x-2)(x+3), so both -3 and 2 work for values of x.
The next part is just telling you the difference between an even and an odd function.
For the next part, you're just composing functions. For the first one f(circle)g, just plug the entire g(x) function into the f(x) function. That's all it is. Then do it backwards, plug the f(x) function into the g(x) function.
Hope that helps you mathematically illiterate faggot.
The image of an element of the domain is the element of the co-domain to which it is paired in the relation. Given a general function f, the image of 0 is f(0). You can't even graph an image since an image is a set.
>>604319029 Sure you can, it's a point at the corresponding coordinates on the graph of the equation. It doesn't ask for that, but you can graph the equation and just highlight the point for value 0. Why so butthurt about this anon?
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