Given the function defined on the interval below, find the value of c where [tex]f(c)[/tex] = the average value.
[tex]f(x)=x^2+x-6[/tex]
on the interval [tex][0,4][/tex]
[tex]c=[?][/tex]
Round to the nearest thousandth.
Answer (2)
Calculate the definite integral of f ( x ) = x 2 + x − 6 from 0 to 4 , which is 3 16 .
Divide the definite integral by the interval length ( 4 − 0 = 4 ) to find the average value: 3 16 /4 = 3 4 ≈ 1.333 .
Set f ( c ) = c 2 + c − 6 equal to the average value 3 4 , resulting in the quadratic equation c 2 + c − 3 22 = 0 .
Solve the quadratic equation for c and choose the solution within the interval [ 0 , 4 ] , rounding to the nearest thousandth: 2.254 .
Explanation
Problem Analysis We are given the function f ( x ) = x 2 + x − 6 on the interval [ 0 , 4 ] . Our goal is to find the value c in this interval such that f ( c ) is equal to the average value of f ( x ) on the interval [ 0 , 4 ] .
Average Value Formula First, we need to calculate the average value of f ( x ) on the interval [ 0 , 4 ] . The formula for the average value of a function f ( x ) on the interval [ a , b ] is given by: Average value = b − a 1 ∫ a b f ( x ) d x In our case, a = 0 , b = 4 , and f ( x ) = x 2 + x − 6 .
Calculating the Definite Integral Now, let's compute the definite integral: ∫ 0 4 ( x 2 + x − 6 ) d x We can integrate term by term: ∫ 0 4 x 2 d x + ∫ 0 4 x d x − ∫ 0 4 6 d x = [ 3 x 3 ] 0 4 + [ 2 x 2 ] 0 4 − [ 6 x ] 0 4 = ( 3 4 3 − 3 0 3 ) + ( 2 4 2 − 2 0 2 ) − ( 6 ( 4 ) − 6 ( 0 ) ) = 3 64 + 2 16 − 24 = 3 64 + 8 − 24 = 3 64 − 16 = 3 64 − 48 = 3 16 So, ∫ 0 4 ( x 2 + x − 6 ) d x = 3 16 .
Finding the Average Value Next, we divide the result of the integral by ( 4 − 0 ) = 4 to find the average value: Average value = 4 1 × 3 16 = 12 16 = 3 4 Thus, the average value of f ( x ) on the interval [ 0 , 4 ] is 3 4 ≈ 1.333 .
Setting up the Equation Now, we set f ( c ) = c 2 + c − 6 equal to the average value 3 4 :
c 2 + c − 6 = 3 4 c 2 + c − 6 − 3 4 = 0 c 2 + c − 3 18 − 3 4 = 0 c 2 + c − 3 22 = 0
Solving for c We can use the quadratic formula to solve for c :
c = 2 a − b ± b 2 − 4 a c In our case, a = 1 , b = 1 , and c = − 3 22 .
c = 2 ( 1 ) − 1 ± 1 2 − 4 ( 1 ) ( − 3 22 ) = 2 − 1 ± 1 + 3 88 = 2 − 1 ± 3 3 + 88 = 2 − 1 ± 3 91 c = 2 − 1 ± 3 91 So, we have two possible values for c :
c 1 = 2 − 1 + 3 91 ≈ 2 − 1 + 5.50757 ≈ 2 4.50757 ≈ 2.253785 c 2 = 2 − 1 − 3 91 ≈ 2 − 1 − 5.50757 ≈ 2 − 6.50757 ≈ − 3.253785 Since we are looking for a value of c in the interval [ 0 , 4 ] , we can discard the negative solution c 2 ≈ − 3.254 .
Final Answer The value c 1 ≈ 2.253785 lies within the interval [ 0 , 4 ] . Rounding to the nearest thousandth, we get c ≈ 2.254 .
Conclusion Therefore, the value of c where f ( c ) equals the average value of f ( x ) on the interval [ 0 , 4 ] is approximately 2.254 .
Examples
Imagine you are designing a temperature control system for a chemical reaction. The rate of reaction is described by the function f ( x ) = x 2 + x − 6 , where x represents the temperature. To optimize the reaction, you need to find a specific temperature c at which the reaction rate equals its average rate over the temperature range [ 0 , 4 ] . This problem demonstrates how finding the average value of a function and solving for a specific point can be crucial in optimizing real-world processes.
To find c such that f ( c ) equals the average value of the function on the interval [ 0 , 4 ] , we first calculate the average value to be 3 4 . Setting f ( c ) = 3 4 , we find a solution using the quadratic formula, resulting in c ≈ 2.254 within the interval. This means c is the point where the function value equals its average on the given interval.
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