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Let for real
MATHEMATICS · Multi-part · ISC · Mathematics · Calculus · Calculus
Problem
Let
f
(
x
)
=
e
(
p
+
1
)
x
−
e
x
f(x)=e^{(p+1)x}-e^x
f
(
x
)
=
e
(
p
+
1
)
x
−
e
x
for real
p
>
0
p>0
p
>
0
. Then: (1) The value
x
=
s
p
x=s_p
x
=
s
p
for which
f
(
x
)
f(x)
f
(
x
)
is minimum is (2) Let
g
(
t
)
=
∫
t
t
+
1
f
(
x
)
e
t
−
x
d
x
g(t)=\int_t^{t+1}f(x)e^{t-x}\,dx
g
(
t
)
=
∫
t
t
+
1
f
(
x
)
e
t
−
x
d
x
. The value
t
=
t
p
t=t_p
t
=
t
p
for which
g
(
t
)
g(t)
g
(
t
)
is minimum is (3) Using
1
+
p
2
≤
e
p
−
1
p
≤
1
+
p
2
+
p
2
1+\frac{p}{2}\leq\frac{e^p-1}{p}\leq1+\frac{p}{2}+p^2
1
+
2
p
≤
p
e
p
−
1
≤
1
+
2
p
+
p
2
for
0
<
p
≤
1
0<p\leq1
0
<
p
≤
1
, find
lim
p
→
0
+
(
s
p
−
t
p
)
\lim_{p\to0^+}(s_p-t_p)
lim
p
→
0
+
(
s
p
−
t
p
)
.
Related theory
Limits
Optimization
Maxima and minima
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