Please help me with this Calculus question about Average Value of a Function(video) or nice explanation

Question

I got the avaerage value its 6/5, however I cannot solve these question

Find c such that f_ave = f(c). (Enter solutions from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes. Round the answers to three decimal places.)

Answer 1

Please see video.

Answer 2

You are correct that the mean is 6/5. Evaluate the integral, -3/(1+x^2), at the 2 endpoints and divide by 2.

Since the function is continuous on [0, 2], we are guaranteed at least one solution. Given that f(0) = 0, f(1) = 3/2, and f(2) = 12/25, we know that there are at least 2 solutions (f(1) > 6/5, and f(0) and f(2) are less)

If we calculate f'(x), we will see that there are exactly 2 solutions - the derivative is negative on ( 1/√3, 2) and positive on (0, 1/√3), as f'(x) =

(6(1+x2)^2 - 24x^2(1+x^2))/(1+x^2)^4 =

(6(1+x^2) - 12x^2)/(1+x^2)^3 =

(6 - 18x^2)/(1+x^2)^3

Now, to find where 6x/(1+x^2)^2 = 6/5, I solved using Excel's solver. You could also use Newton's method or some other process. I plotted x and f(x) for x from 0 to 2, and started the solver near the 2 solutions. I got solutions of .21979, which rounds to .220, and 1.20684, which rounds to 1.207.