You are asked in these exercises to determine whether a piecewise-defined function f is differentiable at a value x = x 0 , where f is defined by different formulas on different sides of x 0 . You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8). Theorem. Let f be continuous at x 0 and suppose that lim x → x 0 f ′ x exists. Then f is differentiable at x 0 , and f ′ x 0 = lim x → x 0 f ' x . Let f x = x 3 + 1 16 , x < 1 2 3 4 x 2 , x ≥ 1 2 Determine whether f is differentiable at x = 1 2 . If so, find the value of the derivative there.
You are asked in these exercises to determine whether a piecewise-defined function f is differentiable at a value x = x 0 , where f is defined by different formulas on different sides of x 0 . You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8). Theorem. Let f be continuous at x 0 and suppose that lim x → x 0 f ′ x exists. Then f is differentiable at x 0 , and f ′ x 0 = lim x → x 0 f ' x . Let f x = x 3 + 1 16 , x < 1 2 3 4 x 2 , x ≥ 1 2 Determine whether f is differentiable at x = 1 2 . If so, find the value of the derivative there.
You are asked in these exercises to determine whether a piecewise-defined function
f
is differentiable at a value
x
=
x
0
, where
f
is defined by different formulas on different sides of
x
0
. You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8). Theorem. Let
f
be continuous at
x
0
and suppose that
lim
x
→
x
0
f
′
x
exists. Then
f
is differentiable at
x
0
, and
f
′
x
0
=
lim
x
→
x
0
f
'
x
.
Let
f
x
=
x
3
+
1
16
,
x
<
1
2
3
4
x
2
,
x
≥
1
2
Determine whether
f
is differentiable at
x
=
1
2
. If so, find the value of the derivative there.
Definition Definition Group of one or more functions defined at different and non-overlapping domains. The rule of a piecewise function is different for different pieces or portions of the domain.
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