Solve for v.h = vt - 16t^2What is v equal to?
1 answer:
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Answer:
A
Step-by-step explanation:
We are given the function f and its derivative, given by:
Remember that f(x) is decreasing when f'(x) < 0.
And f(x) is increasing when f'(x) > 0.
Firstly, determining our zeros for f'(x), we see that:
Since a is a (non-zero) positive constant, -a is negative.
We can create the following number line:
<-----(-a)-----0-----(a)----->
Next, we will test values to the left of -a by using (-a - 1). So:
Since a is a positive constant, (2a + 1) will be positive as well.
So, since f'(x) > 0 for x < -a, f(x) increases for all x < -a.
To test values between -a and a, we can use 0. Hence:
This will always be negative.
So, since f'(x) < 0 for -a < x < a, f(x) decreases for all -a < x < a.
Lasting, we can test all values greater than a by using (a + 1). So:
Again, since a > 0, (2a + 1) will always be positive.
So, since f'(x) > 0 for x > a, f(x) increases for all x > a.
The answer choices ask for the domain for which f(x) is decreasing.
f(x) is decreasing for -a < x < a since f'(x) < 0 for -a < x < a.
So, the correct answer is A.