Find, if it exists, a value c in the interval [1, 4] such that the instantaneous rate of change of f(x) = 12
x.z-dn.net/?f=%5Csqrt%7Bx%7D" id="TexFormula1" title="\sqrt{x}" alt="\sqrt{x}" align="absmiddle" class="latex-formula"> at c is the same as the average rate of change of f over the interval [1, 4]. (If an answer does not exist, enter DNE.)
2 answers:
Answer:
c = 2.25
Step-by-step explanation:
f(x) is a continuous and differentiable function on the interval, so the Mean Value Theorem guarantees a value for c exists.
The average slope is ...
m = (f(4) -f(1))/(4 -1) = (24 -12)/3 = 4
The point at which the derivative is 4 is ...
f'(c) = 6/√c = 4
√c = 6/4
c = 2.25
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Answer: d
Step-by-step explanation:
Find the area of the rectangle.
Find the area of the base of the semicircle.
12 in would be the diameter, divide by 2 to get the radius.
Plug this into the formula.
Since we do not have a full circle but a semi one instead, we must divide our result by 2.
Add both areas.
180+56.52=236.52