Answer:
THE FUNCTION IS ALWAYS INCREASING.
Complete question: The graph of f(x)= power of 3√x + 8 is shown.
Step-by-step explanation:
See attachment for the missing graph.
when x = -16 ; y = -2
when x = -12 ; y = -1.5
when x = -8 ; y = 0
when x = -6 ; y = 1.25
when x = 0 ; y = 2
when x = 6 ; y = 2.5
As the value of x increases, so does the value of y. Thus, the function is always increasing.
Answer:
10%
Step-by-step explanation:
The total population of students will be equal to the sum of both those who wore the silly hat and those who didn't hence
Total students= 74+666=740
Number of students who wore silly hat is given as 74 hence when expressed as a percentage of the total students' population we get that
percentage of the students wore a silly hat=
Therefore, 10% of students wore a silly hat last Tuesday
Answer:
x=126
Step-by-step explanation:
Answer:
<u>-4</u>
3
Step-by-step explanation:
-6y=8x-4
0=8x+6y-4---(i)
slope (m1)=<u>-coefficient</u><u> </u><u>of</u><u> </u><u>x</u>
coefficient of y
=<u>-8</u>
6
= <u>-4</u>
3
As the lines are parallel,
m1=m2
so,
The slope of line parallel to the line having equation -6y=8x-4 is <u>-4</u>
3
To prove this inequality we need to consider three cases. We need to see that the equation is symmetric and that switching the variables x and y does not change the equation.
Case 1: x >= 1, y >= 1
It is obvious that
x^y >= 1, y^x >= 1
x^y + y^x >= 2 > 1
x^y + y^x > 1
Case 2: x >= 1, 0 < y < 1
Considering the following sub-cases:
- x = 1, x^y = 1
- x > 1,
Let x = 1 + n, where n > 0
x^y = (1 + n)^y = f_n(y)
By Taylor Expansion of f_e(y) around y = 0,
x^y = f_n(0) + f_n'(0)/1!*y + f_n''(0)/2!*y^2 + ...
= 1 + ln(1 + n)/1!*y + ln(1 + n)^2/2!*y^2 + ...
Since ln(1 + n) > 0,
x^y > 1
Thus, we can say that x^y >= 1, and since y^x > 0.
x^y + y^x > 1
By symmetry, 0 < x < 1, y >= 1, also yields the same.
Case 3: 0 < x, y < 1
We can prove this case by fixing one variable at a time and by invoking symmetry to prove the relation.
Fixing the variable y, we can set the expression as a function,
f(x) = x^y + y^x
f'(x) = y*x^(y-1) + y^x*ln y
For all x > 0 and y > 0, it is obvious that
f'(x) > 0.
Hence, the function f(x) is increasing and hence the function f(x) would be at its minimum when x -> 0+ (this means close to zero but always greater than zero).
lim x->0+ f(x) = 0^y + y^0 = 0 + 1 = 1
Thus, this tells us that
f(x) > 1.
Fixing variable y, by symmetry also yields the same result: f(x) > 1.
Hence, when x and y are varying, f(x) > 1 must also hold true.
Thus, x^y + y^x > 1.
We have exhausted all the possible cases and shown that the relation holds true for all cases. Therefore,
<span> x^y + y^x > 1
----------------------------------------------------
I have to give credit to my colleague, Mikhael Glen Lataza for the wonderful solution.
I hope it has come to your help.
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