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For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instantaneous rate of change of y with respect to x at the value x=3. Average Rate of Change : Instantaneous Rate of Change at x=3:

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Answer:

The instantaneous rate of change of
y with respect to
x at the value
x = 3 is 18.

Explanation:

a) Geometrically speaking, the average rate of change of
y with respect to
x over the interval by definition of secant line:


r = (y(b) -y(a))/(b-a) (1)

Where:


a,
b - Lower and upper bounds of the interval.


y(a),
y(b) - Function exaluated at lower and upper bounds of the interval.

If we know that
y = 3\cdot x^(2),
a = 3 and
b = 6, then the average rate of change of
y with respect to
x over the interval is:


r = (3\cdot (6)^(2)-3\cdot (3)^(2))/(6-3)


r = 27

The average rate of change of
y with respect to
x over the interval
[3,6] is 27.

b) The instantaneous rate of change can be determined by the following definition:


y' = \lim_(h \to 0)(y(x+h)-y(x))/(h) (2)

Where:


h - Change rate.


y(x),
y(x+h) - Function evaluated at
x and
x+h.

If we know that
x = 3 and
y = 3\cdot x^(2), then the instantaneous rate of change of
y with respect to
x is:


y' = \lim_(h \to 0) (3\cdot (x+h)^(2)-3\cdot x^(2))/(h)


y' = 3\cdot \lim_(h \to 0) ((x+h)^(2)-x^(2))/(h)


y' = 3\cdot \lim_(h \to 0) (2\cdot h\cdot x +h^(2))/(h)


y' = 6\cdot \lim_(h \to 0) x +3\cdot \lim_(h \to 0) h


y' = 6\cdot x


y' = 6\cdot (3)


y' = 18

The instantaneous rate of change of
y with respect to
x at the value
x = 3 is 18.

User Charles Graham
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