Suppose the derivative of a function f is f '(x) = (x + 2)4(x − 3)3(x − 5)6. on what interval is f increasing? (enter your answer in interval notation.)

(3, 5) U (5, ∞)

Step-by-step explanation:

The derivative has zeros at -2, 3, and 5. It changes sign only at x=3, being zero or negative to the left of that point, and being zero or positive to the right of that point. The function increases where the derivative is positive (not zero), so on the interval ...

(3, 5) U (5, ∞)

The function f'(x) will touch the x-axis, but not cross, at x=-2 and x=5, where the powers of the factors (x+2) and (x-5) are even. The function will cross the axis where the power of the factor is odd, at x=3.

Each place where the derivative is zero corresponds to a flat spot in the function. (The function is neither increasing nor decreasing there.) The higher the power of the corresponding factor in the derivative, the "flatter" the function is at that point.

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

she gotta workk that a 2

step-by-step explanation: