h(x) = x+cos(ax) is a family of functions where a is a positive constant such that 0<a<4. Which values of a make h(x) strictly decreasing? Justify your answer.
I can't figure this out and help would be appreciated. Thanks in advance!
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To determine which values of a make h(x) strictly decreasing, we need to analyze the derivative of h(x). A function is strictly decreasing on an interval if its derivative is negative on that interval.
Let's start by finding the derivative of h(x):
h'(x) = 1 - a*sin(ax)
Now, for h(x) to be strictly decreasing, h'(x) must be negative for all values of x within the domain. Recall that the range of the sine function is between -1 and 1. Therefore, a*sin(ax) can take values between -a and a.
To ensure h'(x) is negative for all x, we require -a*sin(ax) to be less than 1 for all x. In other words, we need:
-a*sin(ax) < 1
Dividing both sides of the inequality by -a (since a is positive) gives:
sin(ax) > -1/a
Now, let's consider the range of values for x. In this case, x can be any real number.
To guarantee sin(ax) > -1/a for all x, we need the right side of the inequality, -1/a, to be within the range of the sine function. Since -1 ≤ sin(ax) ≤ 1 for all real numbers x, we should have:
-1 ≤ -1/a ≤ 1
To ensure that the inequality holds, -1/a should be within this range. Hence:
-1 ≤ -1/a ≤ 1
By multiplying both sides of the inequality by -1, we have:
1/a ≤ 1
Finally, by taking the reciprocal of both sides, we get:
a ≥ 1
In conclusion, for h(x) to be strictly decreasing, the values of a should be greater than or equal to 1. Therefore, the valid range for a is 1 ≤ a < 4.
Note: We took the reciprocal of both sides at the end because we initially assumed a < 4. Taking the reciprocal of a ≤ 1 would result in a > 1, which is not valid since a cannot be greater than 4. This is why we obtained a ≥ 1.