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If we stay with positive ∆x,then
∆y = f(1+∆x)−f(1) = 2(1+∆x)−1−1=2∆x,
∆y
∆x
=2
∆x→0
−−−→ 2.
If ∆x<0,
∆y = f(1 + ∆x) − f(1) = 1 + ∆x − 1=∆x,
∆y
∆x
=1
∆x→0
−−−→ 1
We have proven that the derivative at 1 does not exist. However,
one-sided lim
x→1+0
f(x), lim
x→1−0
f(x) limits are both equal to the value
(being 1)off at x =1, which means continuity.
Example 11
One other example is provided by the function y =
3
√
x at x =0.
It turns out that the limit of the difference quotient at x =0is ∞,
which means our function is not differentiable at 0. Clearly, it is
continuous everywhere.
4.5. Other Rates of Change
Suppose y is a quantity that depends on x : y = f(x).Ifx changes
from x
1
to x
2
, then the change in x (or the increment of x)is∆x =
x
2
− x
1
and the corresponding change in y is ∆y = f(x
2
) − f(x
1
).
The difference quotient
∆y
∆x
=
f(x
2
) − f(x
1
)
x
2
− x
1
is called the average rate of y with respect to x over the interval
[x
1
,x
2
] and can be interpreted as the slope of the respective secant
line (if we graph the function).
By analogy with velocity, we consider the average rate of change
over smaller and smaller intervals by letting x
2
approach x
1
and
therefore letting ∆x approach 0.
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