Asked by tt
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A curve labeled b of x and five points on the curve are plotted on a coordinate plane. The curve begins in quadrant 3 sloping upward from left to right, passes through quadrant 2, and then rises through quadrant 1.
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Compare the estimated average rate of change for the graphed cubic function b(x)=3x+9−−−−−√3
to the estimated average rate of change of the square root function d(x)=−3x+9−−−−−−−√
over the interval [−12,−3]
. Which comparison is true?
(1 point)
Responses
The estimated average rate of change of d(x)
is greater than the estimated average rate of change of b(x)
because d(x)
is positive over the interval but b(x)
is negative.
The estimated average rate of change of d times x is greater than the estimated average rate of change of b times x because d times x is positive over the interval but b times x is negative.
The estimated average rate of change of b(x)
is greater than the estimated average rate of change of d(x)
, but both rates are positive.
The estimated average rate of change of b times x is greater than the estimated average rate of change of d times x, but both rates are positive.
The estimated average rate of change of b(x)
is greater than the estimated average rate of change of d(x)
because b(x)
is increasing over the interval but d(x)
is decreasing.
The estimated average rate of change of b times x is greater than the estimated average rate of change of d times x because b times x is increasing over the interval but d times x is decreasing.
The estimated average rate of change of d(x)
is greater than the estimated average rate of change of b(x)
, but both rates are negative.
The estimated average rate of change of d times x is greater than the estimated average rate of change of b times x, but both rates are negative.
All Answers 1
Answered by
GPT-5 mini
AI
b(x) is greater. Over [−12, −3]: b(−12)=∛(3(−12)+9)=∛(−27)=−3 and b(−3)=0, so AROC_b = (0−(−3))/9 = 1/3.
d(−12)=√(−3(−12)+9)=√45≈6.708 and d(−3)=√18≈4.243, so AROC_d ≈ (4.243−6.708)/9 ≈ −0.274.
Thus the average rate of change of b(x) is greater because b is increasing on the interval while d is decreasing.
d(−12)=√(−3(−12)+9)=√45≈6.708 and d(−3)=√18≈4.243, so AROC_d ≈ (4.243−6.708)/9 ≈ −0.274.
Thus the average rate of change of b(x) is greater because b is increasing on the interval while d is decreasing.
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