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:
The distance between the points is approximately 6.4
Step-by-step explanation:
The given coordinates of the points are;
(2, -2), and (6, 3)
The distance between two points, 'A', and 'B', on the coordinate plane given their coordinates, (x₁, y₁), and (x₂, y₂) can be found using following formula;
Substituting the known 'x', and 'y', values for the coordinates of the points, we have;
Therefore, the distance between the points, (2, -2), and (6, 3) = √(41) ≈ 6.4.