Find out which of the linear equations below have a negative slope (select all found)
1 answer:
You might be interested in
Answer:
No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R
Step-by-step explanation:
Assuming: the function is in [0,1]
And rewriting it for the sake of clarity:
Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer
1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:
2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]
The limit to the left
g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)
3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R
Because this is the same as to calculate the limit from the left and right side, of g(x).
This is what the Bilateral Theorem says:
Answer:
(x, y, z) = (3, -2, 4)
Step-by-step explanation:
The attachment shows a solution using a graphing calculator where the first equation is solved for z, and that expression is substituted into each of the other two equations. The solution to that system is (x, y) = (3, -2). Using these values in the expression for z, we find ...
z = 5 -(3 +(-2)) = 4
The solution is (x, y, z) = (3, -2, 4).
__
Your graphing calculator and/or online tools can give you the reduced row echelon form of the augmented matrix representing the system:
__
If you're solving by hand, you can do the substitution described above to eliminate z from one equation. You can eliminate z from another equation by adding the last two:
-3x -4y +4(5 -x -y) = 15 ⇒ -7x -8y = -5
(-3x -4y +4z) +(2x -y -4z) = (15) +(-8) ⇒ -x -5y = 7
This pair of equations can be solved for x and y in any of the usual ways. Using the second equation to substitute for x, for example, gives you ...
-7(-5y -7) -8y = -5 ⇒ 27y = -54 ⇒ y = -2
x = -5(-2) -7 = 3
z = 5 -(3) -(-2) = 4
_____
<em>Additional comment</em>
If all that is needed is a solution, we prefer a graphical or matrix approach. These take about the same amount of time. Both require a little additional effort to determine exact values when the solutions are not integers.
