precalculus problem?

Question

let f and g be increasing functions on the same interval. show that f+g is also increasing on the same interval

please help me, i have no idea how to do
thank u!

Answer 1

If f(x) and g(x) are increasing on (a,b), then:

f(b) > f(a) and g(b) > g(a).

By the addition property of inequality:

f(b) + g(b) > f(a) + g(a), so

(f+g)(b) > (f+g)(a)

Therefore, f+g is increasing on the interval (a,b)

Answer 2

f(x +1) >= f(x)
g(x +1) >= g(x)
f(x + 1) + g(x + 1) >= f(x) + g(x)

When adding things the inequality remains the same (i.e. does not flip) so you can add the two functions together and it is still correct to say that since f and g are both increasing, f + g must be increasing too on the same interval.

Answer 3

ok, you have sqrt(x+a million) + a million = sqrt(2x) sq. the two sides components: x +a million + 2sqrt(x+a million) + a million = 2x this is the comparable as 2sqrt(x+a million) = x - 2 sq. the two sides lower back components: 4(x+a million) = x^2 - 4x + 4 this is the comparable as 4x + 4 = x^2 -4x + 4 furnish all of it to a minimum of one section components: x^2 - 8x = 0 this is the comparable as x(x - 8) = 0 subsequently x = 0 and x = 8 are the concepts 8 of direction works mutually as plugging it lower back indoors the equation, yet you're able to have the means to opt to remember that the sq. root of a million ought to be -a million besides as a million mutually as plugging indoors the x=0. i choose this helps.

Answer 4

Both functions are increasing so:
letting epsilon be any positive value such that x+epsilon is in the interval of interest:
f(x+epsilon)-f(x)>0
&
g(x+epsilon)-g(x)>0

adding 2 positive #'s always gives a positive answer so:
f(x+epsilon)+g(x+epsilon)>f(x)+g(x)

Answer 5

Look in your book. Though I don't remember the exact theorem, you will probably find almost this exact same problem actually covered in the text. All you really need is the name/number of the theorem, and if your teacher is really mean, its proof.

Answer 6

Jenh42002's proof is not correct, but the correct proof follows her basic ideas.

If f and g are increasing on a same interval I, then, for every x_1 and x_2 on I such that x_1 < x_2, we have f(x_1) <= f(x_2) and g(x_1) <= g(x_2). Therefore, f(x_1) + g(x_1) <= f(x_2) + g(x_2), proving your assertion.

Answer 7

[a,b] f(a) g(a)
add f(a)+g(a)
,