How do you determine all values of c that satisfy the mean value theorem on the interval [0,1] for #f(x)= x/(x+6)#?

Question

Adam Adkins · Accepted Answer

undefined To determine all values of $ c $ that satisfy the Mean Value Theorem on the interval $[0,1]$ for $ f(x) = \frac{x}{x+6} $, follow these steps:

1. Calculate the average rate of change of $ f(x) $ on the interval $[0,1]$ using the formula: $ \frac{f(1) - f(0)}{1 - 0} $.

2. Compute the derivative of $ f(x) $ using the quotient rule: $ f'(x) = \frac{(x+6) - x}{(x+6)^2} $.

3. Find all values of $ c $ in the interval $[0,1]$ such that $ f'(c) $ equals the average rate of change calculated in step 1.

4. Substitute the values of $ c $ found in step 3 into the function $ f(x) $ to obtain the corresponding $ f(c) $ values.

The values of $ c $ that satisfy the Mean Value Theorem on the interval $[0,1]$ for $ f(x) = \frac{x}{x+6} $ are the solutions obtained from step 3.

Isaiah Allen · Answer

#c=6-sqrt7~=6-2.646=3.354 in (0,1)#

The Mean Value Theorem statement :-

Let #f# be function : # f : [a,b] rarr RR#.

#(M_1) : # f is continuous on #[a,b]#,

#(M_2) : # f is differentiable on #(a,b)#.

Then, #EE# some #c in (a,b)# such that #f'(c)=(f(b)-f(a))/(b-a)#.

In our case, we find that #f#, being a rational function

# : f : [0,1]rarrRR : f(x)=x/(x+6)#,

is continuous on #[0,1]# and differentiable on #(0,1)#.

Thus the conds. #(M_1) and (M_2)# are satisfied by #f#.

Hence, by M.V.T., #(f(1)-f(0))/(1-0)=1/7=f'(c), where, c in (0,1)#.

But, #f(x)=x/(x+6)=(x+6-6)/(x+6)=(x+6)/(x+6)-1/(x+6)=1-1/(x+6)#

#rarr f'(x)=0-(-1/(x+6)^2)=1/(x+6)^2#.

Therefore, #f'(c)=1/7, c in (0,1) rArr 1/(c+6)^2=1/7 rArr c+6=+-sqrt7#, or,

#c=-6+-sqrt7#

But, because of #c in (0,1), c=-6-sqrt7<0# is not admissible.

#:. c=6-sqrt7~=6-2.646=3.354 in (0,1)#

Enjoy Maths.!