Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
1
Step-by-step explanation:
I have done this by a Calculator
<3
Red
F(3)=5 // f(3)=2(3)-1 = 5
We have

We subtract 11x on both sides and 7 on both sides

we sum like terms

Then we isolate the x

ANSWER
x=-1
The equation only have one solution
Answer:
234 inches
Step-by-step explanation:
Area= base x height
Height= 18 inches
Base= 13 inches
The height of a parallelogram must be perpendicular to its base, the base of this parallelogram can only be 13 inches and the height is 18 inches. Since the side lengths are not straight lines the height must be found using this method.
The height of a triangle, parallelogram and trapezoid must have a height that is perpendicular to its base. Whereas it can be found by extending the length of the base or straightening the side length that is to be the height creating a right angle.