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Sagot :
To find when a function is increasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is positive. Now test values on all sides of these to find when the function is positive, and therefore increasing.
As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function.
Graph of a polynomial that shows the increasing and decreasing intervals and local maximum and minimum.
Figure 3. The function
f
(
x
)
=
x
3
−
1
2
x
f(x)=x
3
−12x is increasing on
(
−
∞
−
)
∪
(
2
,
∞
)
(−∞,−2)∪
(2,∞) and is decreasing on
(
−
2
2
)
(−2,2).
I don’t know how to solve this but example?
A function
f
f is an increasing function on an open interval if
f
(
b
)
>
f
(
a
)
f(b)>f(a) for any two input values
a
a and
b
b in the given interval where
b
>
a
b>a.
A function
f
f is a decreasing function on an open interval if
f
(
b
)
<
f
(
a
)
f(b)a
a and
b
b in the given interval where
b
>
a
b>a.
A function
f
f has a local maximum at
x
=
b
x=b if there exists an interval
(
a
,
c
)
(a,c) with
a
<
b
<
c
ax
x in the interval
(
a
,
c
)
(a,c),
f
(
x
)
≤
f
(
b
)
f(x)≤f(b). Likewise,
f
f has a local minimum at
x
=
b
x=b if there exists an interval
(
a
,
c
)
(a,c) with
a
<
b
<
c
ax
x in the interval
(
a
,
c
)
(a,c),
f
(
x
)
≥
f
(
b
)
f(x)≥f(b).
Solution
We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from
t
=
1
t=1 to
t
=
3
t=3 and from
t
=
4
t=4 on.
In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval
(
4
,
∞
)
(4,∞)
I’m sorry I couldn’t find the answer
As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function.
Graph of a polynomial that shows the increasing and decreasing intervals and local maximum and minimum.
Figure 3. The function
f
(
x
)
=
x
3
−
1
2
x
f(x)=x
3
−12x is increasing on
(
−
∞
−
)
∪
(
2
,
∞
)
(−∞,−2)∪
(2,∞) and is decreasing on
(
−
2
2
)
(−2,2).
I don’t know how to solve this but example?
A function
f
f is an increasing function on an open interval if
f
(
b
)
>
f
(
a
)
f(b)>f(a) for any two input values
a
a and
b
b in the given interval where
b
>
a
b>a.
A function
f
f is a decreasing function on an open interval if
f
(
b
)
<
f
(
a
)
f(b)a
a and
b
b in the given interval where
b
>
a
b>a.
A function
f
f has a local maximum at
x
=
b
x=b if there exists an interval
(
a
,
c
)
(a,c) with
a
<
b
<
c
ax
x in the interval
(
a
,
c
)
(a,c),
f
(
x
)
≤
f
(
b
)
f(x)≤f(b). Likewise,
f
f has a local minimum at
x
=
b
x=b if there exists an interval
(
a
,
c
)
(a,c) with
a
<
b
<
c
ax
x in the interval
(
a
,
c
)
(a,c),
f
(
x
)
≥
f
(
b
)
f(x)≥f(b).
Solution
We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from
t
=
1
t=1 to
t
=
3
t=3 and from
t
=
4
t=4 on.
In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval
(
4
,
∞
)
(4,∞)
I’m sorry I couldn’t find the answer
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