SOLUTION: f(x)=(4x)sqrt(3-x) a)find the zeroes of the function b)approximate the intervals over which the function is increasing, decreasing, or constant c)determine whether the function

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Question 624594: f(x)=(4x)sqrt(3-x)
a)find the zeroes of the function
b)approximate the intervals over which the function is increasing, decreasing, or constant
c)determine whether the function is even, odd, or neither
I tried it out and for a) (0,4)(3,0) c)neither

Answer by jsmallt9(3759) (Show Source):
You can put this solution on YOUR website!
First, please try to choose an appropriate category for your posts. This problem has nothing to do with Trigonometry. (I have changed the category from Trig to a more appropriate one.)

Your answer for (c) is correct.

For part (a) you misunderstood what "zeroes of a function" means. The zeros of a function are the set of inputs values (x's) that result in a zero function value (y). In other "words", the zeroes of your function are the solution set to:

This is already factored. So the solutions will be the x's that make either factor zero:
or
Solving these we should get:
x = 0 or x = 3
These are the zeroes of the function.

For part (b), I'm not sure how to explain it. If you knew Calculus then you could find the first derivative and then analyze it to see when it is positive (increasing), negative (decreasing) or zero (constant):
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
Before we begin the analysis, let us note that the domain of f is . Now let's look at the fraction. The denominator will be positive for all x's. So the numerator will determine whether the derivative is positive, negative or zero. With some thought we should be able to figure out:
f' > 0 when x < 2 so f increases for all these x values
f' = 0 when x = 2 so f is constant only at this one x value
f' < 0 when 2 < x <=3 (remember the domain!) so f is decreasing for these x values.

If you don't know Calculus, another way to figure this out is to use a graphing calculator with a trace function. Below is a graph of this function:

As you can see, there is a "hump" (relative maximum is the proper term for this) where the graph stops going up, turns and starts going down. Enter this function into your graphing calculator and use the trace function try to find where the "hump" is.

Without Calculus and without a graphing calculator I'm not sure what to suggest. Trial and error? Build a table of values and try to figure out the set(s) of x's where the y's get bigger as the x's get bigger and the set(s) of x's where the y's get smaller as the x's get bigger. (In between these sets is where the function is constant.