Still having trouble with this: Find a constant c between 1 and 9 such that the average value of the function f(x) on the interval [1,9] is equal to f(c). I've been told to split the integral, but what next?

Question

anonymous · Answer

$$f(x)=6x^2-4x+2$$

amistre64 · Answer

find the value of (f(9)-f(1))/(9-1)  i believe to get the average value

this tells us the slope between the points (1,f(1)) and (9,f(9))

amistre64 · Answer

once we know that average value of the fucntion we can define that for the value of the function and solve for an appropriate x - which will be c

anonymous · Answer

So, I don't need to use integrals? This is an integral lesson...

amistre64 · Answer

.... well, it is quite possible that my interpretation of the question is off :)

anonymous · Answer

Should we use $$1 \div (b-a)*\int\limits_{1}^{9}f(x)$$?

anonymous · Answer

That's the average value equation...

anonymous · Answer

oops..1=a and 9=b

amistre64 · Answer

f(1) = 4 ; 

f(9) = 9( 6(9) -4) +2 
      = 9(50) + 2
      = 452

452-4      448      
------ = ---- = 56
  9-1          8  


6x^2 -4x +2 = 56 ; when x= (1 +- sqrt(82))/3

amistre64 · Answer

les see if that pans out :)

anonymous · Answer

Well, neither of those are choices. it's multiple choice.

amistre64 · Answer

$$\frac{1}{8}\int\limits_{1}^{9}6x^2-4x+2\ dx$$
$$\frac{1}{8}(2.9^3-2.9^2+2.9-2.1^3+2.1^2-2.1)$$

$$\frac{1}{8}(2.9^3-162+18 -2)=164$$

anonymous · Answer

That's an answer. So, where does the f(c) come in...can you explain that?

amistre64 · Answer

I spose you want a value of c that makes f(x)=164

anonymous · Answer

Oh! That makes a lot of sense. So, 6c^2-4c+2=164?

amistre64 · Answer

http://www.wolframalpha.com/input/?i=6x%5E2-4x%2B2%3D164

anonymous · Answer

You are the best! Thank you!

amistre64 · Answer

yw

amistre64 · Answer

i still think the question needs to be reworded tho :)