1. Relative extrema of functions are the "tops" or the "bottoms" of the curves (the graphs) representing functions. So they are the points at which the tangent lines are horizontal.
2. In short we are looking for the solutions of f'(x)=0
3. f'(x)=
-9=0
Notice:
so that the square root of this expression exists. So x
1 or x
-1
Also in
notice that the root is always positive and 1/9 is positive, so x must be positive. This second condition, combined with the first one tells us that the solution is in x
1
4.
Square both sides
let
Then the 4th degree equation becomes the quadratic equation
The discriminant D=
is not considered as
so t cannot be negative
x=
Answer:
n = 10
Step-by-step explanation:
From the question,
nC6 = nC4
n!/6!(n-6)! = n!/4!(n-4)!
Simplifying n! out of the equation
⇒ 1/6!(n-6)! = 1/4!(n-4)!
Crossmultiplying
4!(n-4)! = 6!(n-6)!
4!(n-4)! = 6×5×4!(n-6)!
(n-4)! = 30(n-6)!
(n-4)(n-5)(n-6)! = 30(n-6)!
⇒ (n-4)(n-5) = 30
n²-4n-5n+20 = 30
n²-9n+20-30
n²-9n-10 = 0
Solving the quadratic equation,
n-10n+n-10 = 0
(n-10n)+(n-10) = 0
n(n-10)+1(n-10) = 0
(n+1)(n-10) = 0
Either
n+1 = 0
n = -1
or
n-10 = 0
n = 10
n canot be a negative value,
Therefore, n = 10
X= [tex] \sqrt{35} is the answer
-27/-2
13.5 is the answer