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Asked Sep 29, 2019
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Problem 3 in the image Parts 1, 2 , and 3

6x3-5 2x - 10
3r342 6
1- 2
-1
sec
-1
(c) lim tan
(d) lim
2
Problem 3 - Suppose f(x) = x x -1.
3
1. Write the statement of the Intermediate Valua Theorem and explain why f has a root in the interval
0, 1].
2. Suppose A is a constant and g(z) = x3
in the interval 0, 1].
x -1 + Ar(x - 1) (2x - 1). Show that g has at least one
and g 3
3. Calculate g
0, 1. For what values of A do we know for sure that g has three roots in [0, 1|?
Show that if A is large enough, g must have three roots in the interval
C3
1
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6x3-5 2x - 10 3r342 6 1- 2 -1 sec -1 (c) lim tan (d) lim 2 Problem 3 - Suppose f(x) = x x -1. 3 1. Write the statement of the Intermediate Valua Theorem and explain why f has a root in the interval 0, 1]. 2. Suppose A is a constant and g(z) = x3 in the interval 0, 1]. x -1 + Ar(x - 1) (2x - 1). Show that g has at least one and g 3 3. Calculate g 0, 1. For what values of A do we know for sure that g has three roots in [0, 1|? Show that if A is large enough, g must have three roots in the interval C3 1

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Expert Answer

Step 1

The problems are concerned with the application of Intermediate value theorems in different contexts

Step 2

a) Intermediate value theroem: Let f(x) be a continuous function defined on the interval [a ,b]. Then f(x) assumes all the values between f(a) and f(b). In other words, if y is between f(a) and f(b) , then there exists an x in [a,b] with f(x)=y

Step 3

(a) Applying IVT (intermediate value theorem), we see that f(x)=...

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f(x) xxover [0,1] f(x) continuous on[0,1] (as f(x) is a polynomial) Let a 0, b 1 and apply IVT f(a) f(0)- f (b) = f()-1 Now,-1 01.So IVT asserts ExE0,],with f(x) = 0.

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Math

Calculus

Continuity