http://www.ck12.org Chapter 14. Concepts of Calculus
x^8 = 2 x
- Show that there is at least one solution to the following equation.
sinx=x+ 2 - When are you not allowed to use the Intermediate Value Theorem?
Answers: - First rewrite the equation:x^8 − 2 x= 0
Then describe it as a continuous function:f(x) =x^8 − 2 x
This function is continuous because it is the difference of two continuous functions.- f( 0 ) = 08 − 20 = 0 − 1 =− 1
- f( 2 ) = 28 − 22 = 256 − 4 = 252
By the Intermediate Value Theorem, there must exist acsuch thatf(c) =0 because− 1 < 0 <252. The numberc
is one solution to the initial equation.
- Write the equation as a continuous function:f(x) =sinx−x− 2
The function is continuous because it is the sum and difference of continuous functions.- f( 0 ) =sin 0− 0 − 2 =− 2
- f(−π) =sin(−π)+π− 2 = 0 +π− 2 > 0
By the Intermediate Value Theorem, there must exist acsuch thatf(c) =0 because− 2 < 0 <π−2. The numberc
is one solution to the initial equation.
- The Intermediate Value Theorem should not be applied when the function is not continuous over the interval.
Practice
Use the Intermediate Value Theorem to show that each equation has at least one real solution.
- cosx=−x
- ln(x) =e−x+ 1
- 2x^3 − 5 x^2 = 10 x− 5
4.x^3 + 1 =x
5.x^2 =cosx
6.x^5 = 2 x^3 + 2 - 3x^2 + 4 x− 11 = 0
- 5x^4 = 6 x^2 + 1
- 7x^3 − 18 x^2 − 4 x+ 1 = 0
- Show thatf(x) =^22 xx−−^35 has a real root on the interval[ 1 , 2 ].
- Show thatf(x) =^32 xx++^14 has a real root on the interval[− 1 , 0 ].
- True or false: A function has a maximum and a minimum in the closed interval[a,b]; therefore, the function is
continuous. - True or false: A function is continuous over the interval[a,b]; therefore, the function has a maximum and a
minimum in the closed interval.