what is the smallest positive integer a such that the intermediate value theorem guarantees a zero exists between 0 and a?
1 answer:
The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
#SPJ4
You might be interested in
9514 1404 393
Answer:
17/99
Step-by-step explanation:
Replace the digits 23 in your example with the digits 17 and you have your answer:
_____
In general, a 2-digit repeat will have 99 as its denominator. If the digits are a multiple of 3 or 11, then the fraction can be reduced. 17 is prime, so the fraction cannot be reduced.