Calculus Tutorial 2.8.005.

 
Problem Statement

{3,5,7,9}
{3,5,7,9}
eval round(-1,2)
eval round(7 - cos(5),2)

Show that cos(5 x) = 7 x has a solution on the interval [0,1].

Hint: Show that f(x) = 7 x - cos(5 x) has a zero in [0,1].

 
Step 1

To show that cos(5 x) = 7 x has a solution in the interval [0,1], we could try to directly solve for some value of x in the interval [0,1] that satisfies the equation.

However, this is not easily done. We can make this problem easier by noting that finding an x satisfying cos(5 x) = 7 x is equivalent to finding an x that satisfies 7 x - cos(5 x) = 0.

We can therefore set f(x) = 7 x − cos(5 x) and look for zeroes of f(x) in the interval [0, 1]. To help prove the existence of a zero of f(x) in [0, 1], we will use the Intermediate Value Theorem.

Question Sequence

Question 1

The Intermediate Value Theorem states that if f(x) is continuous on a closed interval [a,b] and f(a)f(b), then for every value M between f(a) and f(b), there exists at least one value c (a,b) such that f(c) = .

Incorrect.
Correct.

 
Step 2

Question Sequence

Question 3

To apply the Intermediate Value Theorem to f(x) = 7 x − cos(5 x) on [0,1], evaluate f(0) and f(1). (Round your answer to two decimal places.)

f(0) =

f(1) =

Incorrect.
Correct.

 
Step 3

Question 4

Because the graph of f(x) = 7 x − cos(5 x) is continuous in the interval [0, 1], the Intermediate Value Theorem tells us that f(x) will take on all values between f(0) = -1.00 and f(1) ≈ 6.72. Since f(0) < 0 < f(1) and f(x) is continuous, the Intermediate Value Theorem asserts that
f(x) = 7 x − cos(5 x) have a zero in the interval [0, 1].

Using IVT, does the equation f(x) = cos(5 x) - 7 x have a solution on the interval [0,1]?

A.
B.
Incorrect.
Correct.