Question

Given the function `f(x)=-(-2x+6)^(1/2)`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-2,3] and find the c?

Answer 1

The answer is `c=1.75`

Explanation:

The mean value theorem states that if a function `f(x)` is continous on the interval `[a,b]` and differentiable on the interval `(a,b)`, then there exists a point `c` in the interval such that

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

Here,

`f(x)=-(-2x+6)^(1/2)=-sqrt(-2x+6)`

The domain is `x in (-oo, 3]`

The function is continuous and differentiable on the interval `(-2,3)`

Therefore,

`f'(x)=-1/(2sqrt(-2x+6))*-2=1/sqrt(-2x+6)`

`f'(c)=1/sqrt(-2c+6)`

`f(-2)=-sqrt(4+6)=-sqrt10`

`f(3)=-sqrt(-6+6)=0`

Therefore,

`1/sqrt(-2c+6)=(0-(-sqrt10))/(3-(-2))`

`=>`, `1/sqrt(-2c+6)=sqrt10/5`

`=>`, `sqrt(-2c+6)=5/sqrt10`

Squaring both sides

`=>`, `-2c+6=25/10=2.5`

`2c=6-2.5=3.5`

`c=3.5/2=1.75`