Thread: [SOLVED] Combining functions

Given the graph of a function with a few clear ordered pairs. Call it f(x). Can one use those given points on the graph, to sketch (f+f)(x)? Add y values? Add x values? Both?

Originally Posted by EyesForEars

Given the graph of a function with a few clear ordered pairs. Call it f(x). Can one use those given points on the graph, to sketch (f+f)(x)? Add y values? Add x values? Both?

double the y-values (the second coordinate in each pair). do you see why this works?

Honestly no. I wish I could. The coordinates are (-2,2.5)(-1,2)(0,1.5)(1,1)(3,-1) So when you say double, do you mean multiply? So that last coordinate would be (3,-2)? Also, i don't know if it matters, but just looking at it i can tell you its a piecewise. Its linear for the first four points, then it turns curvelinear from (1,1) to (3,-1)

Originally Posted by EyesForEars

Honestly no. I wish I could. The coordinates are (-2,2.5)(-1,2)(0,1.5)(1,1)(3,-1) So when you say double, do you mean multiply?

yes, double means multiply by two. do you realize that (f + f)(x) = f(x) + f(x) = 2f(x)...or in other words, 2 times the y-value.

So that last coordinate would be (3,-2)?

yes

Also, i don't know if it matters, but just looking at it i can tell you its a piecewise. Its linear for the first four points, then it turns curvelinear from (1,1) to (3,-1)

no, we cannot tell from this. we just have discrete points here. descriptions like "piece-wise" do not apply. if these points are just a few points given for a non-discrete function, then we cannot tell whether the function is piece-wise or not here.

Ah! I think i get it! Tell me if im on the right track. So, if your going to add to itself you would get . Hence,

Originally Posted by EyesForEars

Ah! I think i get it! Tell me if im on the right track. So, if your going to add to itself you would get . Hence,

Exactly! Operations on functions may look bad, but they're really pretty simplistic. The "(f + f)(x)" means nothing more than "f(x) + f(x)", or "2f(x)". So if f(x) = mx + b, then (f + f)(x) = 2f(x) = 2(mx + b) = 2mx + 2b.

Good work!