Question

Докажите что функция f(x)=√x-15 является возрастающей

Answer 1

Найдем область определения
D(x):[15;+бесконечности)
найдем производную
f'(x)=1/2sqrt(x-15)
производная будет строго пложительна,т.к числетель >0 и знаменатель строго >0
т.к производная положительна,то график будет возрастать на интервале (15;+бесконечности)

Answer 2

Proof that the function f(x) = √x - 15 is increasing

To prove that the function f(x) = √x - 15 is increasing, we need to show that for any two values of x, say x1 and x2, if x1 < x2, then f(x1) < f(x2).

Let's consider two arbitrary values, x1 and x2, where x1 < x2. We need to compare f(x1) and f(x2).

Substituting x1 into the function f(x) = √x - 15, we get:

f(x1) = √x1 - 15

Similarly, substituting x2 into the function, we get:

f(x2) = √x2 - 15

Since x1 < x2, we can conclude that √x1 < √x2 because the square root function is increasing.

Therefore, we have:

√x1 < √x2

Adding -15 to both sides of the inequality, we get:

√x1 - 15 < √x2 - 15

This shows that f(x1) < f(x2), which means the function f(x) = √x - 15 is increasing.

Please note that the sources provided did not directly address the proof of the function's increasing nature. However, the proof is based on the properties of the square root function and the comparison of two values.