Question

F(x) = x/x+6, [1, 12] If it satisfies the hypotheses, find all numbers c that satisfies the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).

Answer 1

The answer is "
`{c=-6+3√(14)`"

Explanation:

Given:

`\to F(x) = (x)/(x+6) , \ [1, 12]`

`\to f'(c)=(f(b)-f(a))/(b-a)`

As per the given intervals:

`[1,12]\\\\ a=1 \\ b=12.`

Calculating the values of f(a) and f(b).

`\to f(1)=(1)/(1+6)= (1)/(7) \\\\ \to f(2)=(12)/(12+6)= (12)/(18)=(2)/(3) \\\\ \to f(x)=(x)/(x+6)\\\\`

`\to f\:'\left(x\right)=(d)/(dx)\left((x)/(x+6)\right)\: \\\\`

`=((d)/(dx)\left(x\right)\left(x+6\right)-(d)/(dx)\left(x+6\right)x)/(\left(x+6\right)^2) \\\\=(1\cdot \left(x+6\right)-1\cdot \:x)/(\left(x+6\right)^2)\\\\`

`\to f'(x)=(6)/(\left(x+6\right)^2)`

Calculating the value of
`\ f'(c):`

`\to \mathbf{f\:'\left(c\right)=(6)/(\left(c+6\right)^2)}`

Replacement of the values in the mean theorem of value now.

`\to (6)/(\left(c+6\right)^2)=((2)/(3)-(1)/(7))/(12-1)\\\\\to (6)/(\left(c+6\right)^2)=((11)/(21))/(11)\\\\\to (6)/(\left(c+6\right)^2)=(1)/(21)\\\\`

Apply the cross multiplication:

`\to (c+6)^2=126 \\\\\to c^2+12c +36= 126\\\\\to c^2+12c +36-126= 126-126 \\\\\to c^2+12c -90= 0\\\\\to c=(-12\pm √(12^2-4\cdot \:1\left(-90\right)))/(2\cdot \:1)\\\\ \to \mathbf{c=-6+3√(14)} \\\\`