Firstly, let's discuss what it means if a function is increasing or decreasing in an interval. A function 'f' is said to be increasing on an interval 'I' if for any two numbers 'x' and 'y' in 'I', if x < y, then f(x) < f(y). Similarly, the function 'f' is said to be decreasing on an interval 'I' if for any two numbers 'x' and 'y' in 'I', if x < y, then f(x) > f(y).
The intervals of increase or decrease for a polynomial function can be found by finding its derivative, setting the derivative equal to zero to find the critical points, and then analysing the sign of the derivative in the intervals created by the critical points.
(a) Function: f(x)=x^2+2x-5
The derivative of the function is: f'(x) = 2x + 2.
Setting f'(x) equal to zero gives x = -1 as the critical point. The sign of the derivative changes at x = -1. Therefore, the function is decreasing on the interval (-inf, -1) and increasing on the interval (-1, inf).
(b) Function: f(x)=10-6x-2x^2
The derivative of the function is: f'(x) = -6 - 4x.
Setting f'(x) equal to zero gives x = -3/2 as the critical point. The sign of the derivative changes at x = -3/2. Therefore, the function is increasing on the interval (-inf, -3/2) and decreasing on the interval (-3/2, inf).
(c) Function: f(x)=-2x^3-9x^2-12x+1
The derivative of the function is: f'(x) = -6x^2 - 18x - 12.
Setting f'(x) equal to zero gives x = -2 and x = 1 as the critical points. The sign of the derivative changes at each critical point. Therefore, the function is decreasing on the interval (-inf, -2), increasing on the interval (-2, 1), and decreasing on the interval (1, inf).
(d) Function: f(x)=6-9x-x^2
The derivative of the function is: f'(x) = -9 - 2x.
Setting f'(x) equal to zero gives x = -9/2 as the critical point. The sign of the derivative changes at x = -9/2. Therefore, the function is increasing on the interval (-inf, -9/2) and decreasing on the interval (-9/2, inf).
(e) Function: f(x)=(x+1)^3(x-3)^3
The derivative of the function is: f'(x) = 3(x+1)^2(x-3)^3 + 3(x+1)^3(x-3)^2.
Setting f'(x) equal to zero gives x = -1 and x = 3 as the critical points. The sign of the derivative does not change at either critical point. Therefore, the function is increasing on the interval (-inf, -1), decreasing on the interval (-1, 3), and increasing on the interval (3, inf).