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Question: How do you use a graph to solve an equation on the interval?...

How do you use a graph to solve an equation on the interval?

Explanation

Solution

Hint : The given question is asked how we solve an equation by using a graph on the given interval. This question is asking the general format for solving an equation with the help of a graph. So let us explain the step by process to solve an equation on the interval only by using a graph and also with an example.

Complete step-by-step answer :
To solve an equation by using a graph on the given interval follow the steps given below:
Step 1: For a given equation on interval find the function yy corresponding to the equation.
Step 2: Draw the table for different values of xx.
Step 3: Graph the function yy, for the values on the table above.
Step 4: Observe the graph and note the points for the given equation. Those points are the solution for the given equation.
Let us consider an equation tan(x)=1\tan (x) = 1 on an interval [2π,2π][ - 2\pi ,2\pi ].
Step 1: Find the function yy,
Given that tan(x)=1\tan (x) = 1, then the function y=tan(x)1y = \tan (x) - 1.
Step 2: Draw the table for different values of xx.
On the given interval assign the different values for xx and thus find y=tan(x)1y = \tan (x) - 1.
Given interval [2π,2π][ - 2\pi ,2\pi ], the values between these interval are: 2π,5π4,3π4 - 2\pi ,\dfrac{{ - 5\pi }}{4},\dfrac{{3\pi }}{4} and 2π2\pi .
Now, x=2πy=tan(2π)1x = - 2\pi \Rightarrow y = \tan ( - 2\pi ) - 1
tan(2π)=0\tan ( - 2\pi ) = 0, by substituting this value,
y=01y = 0 - 1
y=1y = - 1
x=5π4y=tan(5π4)1x = \dfrac{{ - 5\pi }}{4} \Rightarrow y = \tan \left( {\dfrac{{ - 5\pi }}{4}} \right) - 1
Substituting the value tan(5π4)=1\tan \left( {\dfrac{{ - 5\pi }}{4}} \right) = - 1,
y=11y = - 1 - 1
y=2y = - 2.
x=3π4y=tan(3π4)1x = \dfrac{{3\pi }}{4} \Rightarrow y = \tan \left( {\dfrac{{3\pi }}{4}} \right) - 1
We know that tan(3π4)=1\tan \left( {\dfrac{{3\pi }}{4}} \right) = - 1,
y=11y = - 1 - 1
y=2y = - 2.
x=2πy=tan(2π)1x = 2\pi \Rightarrow y = \tan (2\pi ) - 1
Substitute tan(2π)=0\tan (2\pi ) = 0,
y=01y = 0 - 1
y=1y = - 1
Now draw the table for the values we found.

xxy=tan(x)1y = \tan (x) - 1(x,y)(x,y)
2π- 2\piy=tan(2π)1y = \tan ( - 2\pi ) - 1(2π,1)( - 2\pi , - 1)
5π4\dfrac{{ - 5\pi }}{4}y=tan(5π4)1y = \tan \left( {\dfrac{{ - 5\pi }}{4}} \right) - 1(5π4,2)\left( {\dfrac{{ - 5\pi }}{4}, - 2} \right)
3π4\dfrac{{3\pi }}{4}y=tan(3π4)1y = \tan \left( {\dfrac{{3\pi }}{4}} \right) - 1(3π4,2)\left( {\dfrac{{3\pi }}{4}, - 2} \right)
2π2\pi y=tan(2π)1y = \tan (2\pi ) - 1(2π,1)(2\pi , - 1)

Step 3: Graph the function y=tan(x)1y = \tan (x) - 1
Draw the coordinate plane and plot the points we found for the table and connect the points.

Step 4: Now, observe the graph, the solutions for tan(x)=1\tan (x) = 1 on interval [2π,2π][ - 2\pi ,2\pi ] are (7π4,3π4,π45π4)\left( {\dfrac{{ - 7\pi }}{4},\dfrac{{ - 3\pi }}{4},\dfrac{\pi }{4}\dfrac{{5\pi }}{4}} \right).
Hence we got the solution of tan(x)=1\tan (x) = 1 on an interval [2π,2π][ - 2\pi ,2\pi ] by using a graph.

Note : To plot a graph it is necessary to have both xx value and yy value to mark on xx-axis and yy- axis. Therefore convert the equation into function to obtain two points. And also be very careful while converting the equation into function. And then note the points in a form of table to identify them easily. The equation can also be solved manually by using a set of equations and formulas.