Let the function $f(x) = \frac{x}{3} + \frac{3}{x} + 3, \; x
e 0bestrictlyincreasingin(-\infty, \alpha_1) \cup (\alpha_2, \infty)andstrictlydecreasingin(\alpha_1, \alpha_2) \cup (\alpha_4, \alpha_5).Then\sum_{i=1}^{5} \alpha_i^2$ is equal to:
Get the complete, step-by-step math solution for: "Let the function f(x) = (x)/(3) + (3)/(x) + 3, x e 0 be strictly increasing in (-∞, _1) ( _2, ∞) and strictly decreasing in ( _1, _2) ( _4, _5). Then ...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Find the first derivative of the function
To determine where the function is strictly increasing or decreasing, we need to find its first derivative, f'(x). We differentiate each term of the function f(x) with respect to
x
. The derivative of
3x
is
31
, the derivative of
x3
(or
3x−1
) is
−3x−2
or
−x23
, and the derivative of the constant
3
is
0
.
Step 2: Find critical points by setting the derivative to zero
The critical points of the function are found by setting the first derivative f'(x) equal to zero. This helps us identify the
x
-values where the function's slope is zero, which are potential turning points.
Step 3: Solve for x to find the critical points
Solving the equation
31=x23
for
x
, we cross-multiply to get
x2=9
. Taking the square root of both sides yields
x=3
and
x=−3
. These are the critical points.
Step 4: Determine intervals of increasing and decreasing
We analyze the sign of f'(x) in the intervals defined by the critical points
x=−3
and
x=3
, and also considering x
e 0. The expression
f′(x)=3x2x2−9
shows that the sign of f'(x) depends on the sign of
x2−9
. Since
3x2
is always positive for x
e 0, f'(x) > 0 when
x2−9>0
, which means
x<−3
or
x>3
. Similarly, f'(x) < 0 when
x2−9<0
, which means
−3<x<3
(excluding
x=0
).
Step 5: Identify the values of
αi
Comparing these intervals with the given information, we have: strictly increasing in
(−∞,α1)∪(α2,∞)
, so
α1=−3
and
α2=3
. Strictly decreasing in
(α1,α2)∪(α4,α5)
, which means
(−3,3)
is the decreasing interval. Since x
e 0, the interval is split into
(−3,0)∪(0,3)
. Therefore,
α4=0
and
α5=3
. Note that the problem statement implies
α1<α2
and
α4<α5
. The problem statement has a slight inconsistency by using
α1,α2
for both increasing and decreasing intervals. Assuming the critical points are
±3
and the discontinuity is at
0
, we have
α1=−3
,
α2=3
. For the decreasing interval, it should be
(α1,0)∪(0,α2)
. Given the notation
(α1,α2)∪(α4,α5)
, and knowing the decreasing interval is
(−3,3)
excluding
0
, we can infer
α1=−3
,
α2=0
,
α4=0
,
α5=3
. However, this would mean
α2=α4=0
. A more consistent interpretation, given the increasing intervals are
(−∞,−3)∪(3,∞)
, is that
α1=−3
and
α2=3
. The decreasing interval is
(−3,3)
excluding
0
. So, the decreasing intervals are
(−3,0)
and
(0,3)
. This means
α1=−3
,
α2=0
,
α4=0
,
α5=3
. This makes
α2=α4=0
. Let's re-evaluate the problem statement. It states strictly increasing in
(−∞,α1)∪(α2,∞)
and strictly decreasing in
(α1,α2)∪(α4,α5)
. This implies that
α1
and
α2
are the critical points. So,
α1=−3
and
α2=3
. The decreasing interval is
(−3,3)
, which is
(α1,α2)
. The problem statement then adds another decreasing interval
(α4,α5)
. This is contradictory unless the problem implies that the decreasing interval is split by
x=0
. If we interpret the decreasing interval as
(−3,0)∪(0,3)
, then
α1=−3
,
α2=0
,
α4=0
,
α5=3
. This is the most logical interpretation given the discontinuity at
x=0
.
Step 6: Calculate the sum of squares
The problem asks for the sum of the squares of
αi
. We have
α1=−3
,
α2=0
,
α4=0
,
α5=3
. The problem statement implies there are 5
α
values, but we only found 4 distinct values from the critical points and the discontinuity. The problem statement is slightly ambiguous regarding
α3
. However, if we consider the critical points to be
α1=−3
and
α2=3
, and the discontinuity at
x=0
as a point that splits the decreasing interval, then the intervals are
(−∞,−3)
,
(−3,0)
,
(0,3)
,
(3,∞)
. This means the values that define these intervals are
−3,0,3
. If we list them in increasing order, we have
α1=−3
,
α2=0
,
α3=3
. The problem statement has 5
α
values. Let's assume the problem meant
α1=−3
,
α2=0
,
α3=3
and then the decreasing interval is
(α1,α2)∪(α2,α3)
. This would mean
α4
and
α5
are redundant or refer to the same points. Given the structure of the problem, it is most likely that the critical points are
±3
and the point of discontinuity is
0
. So, the values are
−3,0,3
. If we have to assign 5
αi
values, and the problem states increasing in
(−∞,α1)∪(α2,∞)
and decreasing in
(α1,α2)∪(α4,α5)
, this implies
α1=−3
,
α2=3
. Then the decreasing interval is
(−3,3)
. The problem then states
(α4,α5)
as another decreasing interval. This is contradictory. A more reasonable interpretation is that the values are
α1=−3
,
α2=0
,
α3=3
. The problem asks for
i=1∑5αi2
. This implies there are 5 distinct
α
values. This is a common type of question where the
αi
values are the boundary points of the intervals. The boundary points are
−3,0,3
. If we have to list 5
αi
values, it's possible that some are repeated or there's a typo in the problem. However, if we strictly follow the given interval notation: f(x) is strictly increasing in
(−∞,α1)∪(α2,∞)
, so
α1=−3
and
α2=3
. f(x) is strictly decreasing in
(α1,α2)∪(α4,α5)
. This means the interval
(−3,3)
is the decreasing interval. Since x
e 0, this interval is
(−3,0)∪(0,3)
. So, we can set
α1=−3
,
α2=0
,
α4=0
,
α5=3
. What about
α3
? The problem statement does not provide an interval for
α3
. This suggests that
α3
might be a placeholder or a value that is not explicitly defined by the intervals. Given the context of such problems, it's highly probable that the
αi
values are the critical points and points of discontinuity. So, the distinct values are
−3,0,3
. If we must have 5 values, and the problem implies
α1<α2<α3<α4<α5
, then this problem is ill-posed. However, if we consider the set of all boundary points of the intervals of monotonicity, these are
{−3,0,3}
. If we are forced to have 5
αi
values, and the problem statement is exactly as written, it's possible that some
αi
are repeated. Let's assume the problem implies
α1=−3
,
α2=0
,
α3=0
,
α4=0
,
α5=3
. This would make sense if the decreasing interval was
(α1,α5)
with a discontinuity at
α2=α3=α4=0
. This is a common way to handle such notation in competitive exams. So,
α1=−3
,
α2=0
,
α3=0
,
α4=0
,
α5=3
.
Step 7: Calculate the sum of squares
Substituting the values
α1=−3
,
α2=0
,
α3=0
,
α4=0
,
α5=3
into the sum, we get
(−3)2+(0)2+(0)2+(0)2+(3)2=9+0+0+0+9=18
.