Question

Use the intermediate value theorem to prove that the equation

ln? = ? − square root(?) has atleast one solution between ?=2 and ?=3

Answer #1

Use the Intermediate Value Theorem and the Mean Value Theorem to
prove that the equation cos (x) = -2x has exactly one real
root.

for the equation f(x) = e^x - cos(x) + 2x - 3
Use Intermediate Value Theorem to show there is at least one
solution.
Then use Mean Value Theorem to show there is at MOST one
solution

Use the rational root theorem to argue square root of 2 plus
cube root of 3 is irrational

Use the intermediate value theorem (only) to prove that any real
polynomial of odd degree has at least one real solution. Is the
same conclusion true if the degree is even?

Let g (x) = 2^-x −x − x^3, defined on [0,1]
(a) Use the Intermediate Value Theorem (TVI) to prove that the
equation g (x) = 0 has a solution.
(b) Use the Average Value Theorem (TVM) to show that the solution
in part (a) is the only one that exists.

(i) Use the Intermediate Value Theorem to prove that there is a
number c such that 0 < c < 1 and cos (sqrt c) = e^c- 2.
(ii) Let f be any continuous function with domain [0; 1] such
that 0smaller than and equal to f(x) smaller than and equal to 1
for all x in the domain. Use the Intermediate Value Theorem to
explain why there must be a number c in [0; 1] such that f(c)
=c

1. Use the Intermediate Value Theorem to show that
f(x)=x3+4x2-10 has a real root in the
interval [1,2]. Then, preform two steps of Bisection method with
this interval to find P2.

x^5 +x^3 +x +1=0
use the IVT and Rolle's theorem to prove that the equation has
exactly one real solution.

In the style of the proof that square root of 2 is irrational,
prove that the square root of 3 is irrational. Remember, we used a
proof by contradiction. You may use the result of Part 1 as a
"Lemma" in your proof.

If g (x) = 5 + x − 8cot?(pix/8)?, defined on [2,4]
(a) Use the Intermediate Value Theorem (TVI) to prove that the
equation g (x) = 0 has a solution.
(b) Use the Average Value Theorem (TVM) to show that the
solution in part (a) is the only one that exists.

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