Question

Найти промежутки возрастания и убывания функции: 1) y=x^2+6x+25 2) y=x^3-9x+125 3) y=ln x С пошаговым решением, пожалуйста!(..

Answer 1

Ответ:в прикреплённом файле

Пошаговое объяснение:

Answer 2

Промежутки возрастания и убывания функций

1) y=x^2+6x+25 To find the intervals of increase and decrease for the function y=x^2+6x+25, we can start by finding its derivative and then identifying the critical points.

The derivative of y=x^2+6x+25 is y'=2x+6. To find the critical points, we set y' equal to 0 and solve for x: 2x+6=0 2x=-6 x=-3

Now, we can use the first derivative test to determine the intervals of increase and decrease. When x<-3, y' is negative, indicating that the function is decreasing. When x>-3, y' is positive, indicating that the function is increasing.

So, the function y=x^2+6x+25 is decreasing for x < -3 and increasing for x > -3.

2) y=x^3-9x+125 For the function y=x^3-9x+125, we'll follow the same process. The derivative of y=x^3-9x+125 is y'=3x^2-9. Setting y' equal to 0 and solving for x gives us the critical points: 3x^2-9=0 3(x^2-3)=0 x^2-3=0 x^2=3 x=±√3

Using the first derivative test, we can determine the intervals of increase and decrease. When x< -√3, y' is negative, indicating that the function is decreasing. When -√3 < x < √3, y' is positive, indicating that the function is increasing. When x > √3, y' is positive, indicating that the function is increasing.

So, the function y=x^3-9x+125 is decreasing for x < -√3, increasing for -√3 < x < √3, and increasing for x > √3.

3) y=ln(x) For the function y=ln(x), we'll approach it slightly differently. The derivative of y=ln(x) is y'=1/x. The critical points occur where y' is equal to 0 or undefined, which is at x=0.

Using the first derivative test, we can determine that the function y=ln(x) is increasing for x > 0 and undefined for x < 0.

So, the function y=ln(x) is increasing for x > 0.

I hope this helps! Let me know if there's anything else you'd like to discuss or any other questions you have.