366cee…: “As per a tall angle”

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1 :

Worked Out Examples Problem Videos

The following videos may be helpful when trying to solve the problems in this practice section. Note that you may skip to the end of the video to get completion credit for this page if you don’t need to watch them.

Practice Problems: Intermediate Value Theorem (IVT)

Use the Intermediate Theorem to show that the equation $x^2=\sqrt {x+1}$ has a root (a solution) in the interval $(1,2)$ .

Let’s solve the problem in steps.

REMARK: Notice that the equation above is equivalent to the equation $x^2-\sqrt {x+1}=0$ STEP 1

In order to use the Intermediate Value Theorem, we have to define a suitable function.

Let $f(x)=x^2-\sqrt {x+1}$ where

$\text {Domain of } f=[\answer {1},\answer {2}].$ Then $f$ is continuous on its domain.

STEP 2

1.1 :

Find the values. $f(1)=\answer {1-\sqrt {2}}, f(2)=\answer {4-\sqrt {3}}.$

1.2 : Let $L=\answer {0}$ Select all the following statements that are correct.

$f(1)< L$ $f(1)> L$ $f(2)< L$ $f(2)> L$

1.3 : Select all the following statements that are correct.

Since $f(2)< L<f(1)$ , the IVT implies that there is a number $c$ in $\left (1,2\right )$ such that $f(c)=L$ . Since $f(1)< L<f(2)$ , the IVT implies that there is a number $c$ in $\left (f(1),f(2)\right )$ such that $f(c)=L$ . Since $1< L< 2$ , the IVT implies that there is a number $c$ in $\left (1,2\right )$ such that $f(c)=L$ . Since $f(1)< L< f(2)$ , the IVT implies that there is a number $c$ in $\left (1,2\right )$ such that $f(c)=L$ .

When we apply the Intermediate Value Theorem to $f$ on the interval $\left [1,2\right ]$ with $L=0$ , we are guaranteed the existence of at least one point $c$ such that. $f(c)=\answer {0}$ The number $c$ is a solution, or a root, of the original equation.