A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.
Let's look at your function.


is in the numerator. It is defined for every value of x. There is no problem there.

is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.
We set it equal to zero and solve the equation for x.



For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is:
Take the vector u = <ux, uy> = <4, 3>.
Find the magnitude of u:
||u|| = sqrt[ (ux)^2 + (uy)^2]
||u|| = sqrt[ 4^2 + 3^2 ]
||u|| = sqrt[ 16 + 9 ]
||u|| = sqrt[ 25 ]
||u|| = 5
To find the unit vector in the direction of u, and also with the same sign, just divide each coordinate of u by ||u||. So the vector you are looking for is
u/||u||
u * (1/||u||)
= <4, 3> * (1/5)
= <4/5, 3/5>
and there it is.
Writing it in component form:
= (4/5) * i + (3/5) * j
I hope this helps. =)
Answer:
1; -11+ 26 =15
2: 2- (-3) =5
3: 10+ (-4) = 7
4: 11 + -8 =3
5: -2 + 3 =1
6: 8 - 12 = -4
Step-by-step explanation:
!
Answer:
A) to determine the slope for the graph of F(x)
Step-by-step explanation: