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:
Step-by-step explanation:
Substituting x and y values into the equations, we can get the equation
y = 7 -3x
slope: -3
180 miles? seems like the question is incomplete
Using Pythagorean theorem, the student walked 53.58 meters more compared to the total displacement from the starting point.
If a student walks 100 meters north, then 100 meters west, then the path he travels resembles the sides of a right triangle (see attached photo).
Using Pythagorean theorem, we can solve for the total displacement from the starting point to the end point.
c^2 = a^2 + b^2
where c is the total displacement from the starting point to the end point
a is the distance he walks up north
b is the distance he walks to the west
c^2 = 100^2 + 100^2
c^2 = 10,000 + 10,000
c^2 = 20,000
c = 141.42 meters
Comparing the total distance the student walked and the total displacement from the starting point to the end point by subtraction.
100 meters + 100 meters - 141.42 meters = 53.58 meters
Learn more about Pythagorean Theorem here: brainly.com/question/343682
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