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Mathematics
Mathematics
Problem
Consider the function
f
(
x
)
=
{
2
x
,
x
≤
1
2
,
x
>
1
f(x)=\begin{cases}2x, & x\le 1\\2,& x>1\end{cases}
f
(
x
)
=
{
2
x
,
2
,
x
≤
1
x
>
1
and the piecewise-defined function
F
(
x
)
=
{
x
2
+
a
,
x
≤
1
2
x
+
b
,
x
>
1
F(x)=\begin{cases}x^{2}+a, & x\le 1\\2x+b,& x>1\end{cases}
F
(
x
)
=
{
x
2
+
a
,
2
x
+
b
,
x
≤
1
x
>
1
, where
a
a
a
and
b
b
b
are real constants. Find the values of
a
a
a
and
b
b
b
such that
F
F
F
is an antiderivative of
f
f
f
on
R
\mathbb{R}
R
and
F
(
1
)
=
4
F(1)=4
F
(
1
)
=
4
.
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