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1 answer:
Answer:
(A) -3 ≤ x ≤ 1
Step-by-step explanation:
The given function is presented as follows;
h(x) = x² - 1
From the given function, the coefficient of the quadratic term is positive, and therefore, the function is U shaped and has a minimum value, with the slope on the interval to the left of <em>h</em> having a negative rate of change;
The minimum value of h(x) is found as follows;
At the minimum of h(x), h'(x) = d(h(x)/dx = d(x² - 1)/dx = 2·x = 0
∴ x = 0/2 = 0 at the minimum
Therefore, the function is symmetrical about the point where x = 0
The average rate of change over an interval is given by the change in 'y' and x-values over the end-point in the interval, which is the slope of a straight line drawn between the points
The average rate of change will be negative where the y-value of the left boundary of the interval is higher than the y-value of the right boundary of the interval, such that the line formed by joining the endpoints of the interval slope downwards from left to right
The distance from the x-value of left boundary of the interval that would have a negative slope from x = 0 will be more than the distance of the x-value of the right boundary of the interval
Therefore, the interval over which <em>h</em> has a negative rate of change is -3 ≤ x ≤ 1
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Answer:
Step-by-step explanation:
The given functions are and
.
We now composed the two functions to find:
This function is defined if the denominator is not zero.
We write this in interval notation as:
We need to be cautious here as x=0 is not in the domain of g(x).
Therefore the domain of is