Question

Question #b60d4

Answer 1

Please see below.

Explanation:

H1: `f` is continuous on `[1,16]` is true because `f` is continuous on its domain and its domain is all reals except `-8`. And `-8` is not in [1,16]#.

H2: `f` is differentiable on `(1,16)` is true because `f'(x) = 8/(x+8)^2` is defined for all `x` except `-8` and `-8` is not in (1,16)#.

To find `c`, solve `8/(x+8)^2 = (f(16)-f(1))/(16-1)`.

This is `8/(x+8)^2 =1/27`

So `8*27 = (x+8)^2`

And `(x+8)^2 = 8*27` if and only if

`x+8 = +-sqrt(8*27) = +-6sqrt6`

`x = -8+-6sqrt6`

Note that `2 < sqrt6 < 3`,

so `-8+6sqrt6` is between `-8+12 = 4` and `-8 + 18 = 10`

so `-8+6sqrt6` is in `(1,16)`.

Answer 2

The value of `c=6.7`

Explanation:

The mean value theorem states : If `f` is is a function defined and continuous on the Inteval `I=[a,b]` and differentiable on the interval `(a,b)`, then there is a value `c in (a,b)` such that

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

Here,

`f(x)=x/(x+8)` is defined and continuous on the interval `[1,16]`

`f(1)=1/9`

`f(16)=16/24=2/3`

`f'(x)=(1*(x+8)-x*1)/(x+8)^2=8/(x+8)^2`

`f(x)` is differentiable on the interval `(1,16)`

`f'(c)=8/(c+8)^2=(f(16)-f(1))/(16-1)=(6/9-1/9)/(15)=5/9*1/15=1/27`

`(c+8)^2=27*8=216`

`c^2+16c+64-216=0`

`c^2+16c-152=0`

This is a quadratic equation in `c`, the determinant is

`Delta=b^2-4ac=16^2+4*152=864`

As `Delta>0`, there are 2 real solutions

`c=(-16+-sqrt864)/2=(-16+-29.4)/2`

`c_1=6.7`

`c_2=-22.7`

`{c_1} in [1,16]`