If ƒ(x) = -5n + 1 and g(x) = -6n + 2, find (ƒ + g)(-2).

Mathematics

1 answer:

Stels [109]3 years ago

3 0

(f+g) (n) = (–5n +1 ) + (- 6n +2) = –11n +3

(f+g) (n) = –11n +3

(f+g) (–2) = – 11 (–2) +3 = 22 +3 = 25

I hope I helped you^_^

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Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$