Question

How do you determine whether the Mean Value Theorem can be applied to $f(x) = \sqrt {x - 7} $ on the interval $\left[ {11,23} \right]$?

Answer 1

Hint: Here we have to check the conditions of the Mean value theorem by using the given data.
First we have to check if the given function is continuous on the given interval or not and also check if it is differentiable or not. After some simplification we get the required answer.

Complete step-by-step solution:
Given $f(x) = \sqrt {x - 7} $ on the interval $\left[ {11,23} \right]$
The Mean Value Theorem has two hypotheses:
$f$ is continuous on the closed interval $\left[ {a,b} \right]$
$f$ is differentiable open interval $(a,b)$
Then there exist at least one point $x = c$ in the interval, such that
$f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$
In this question, $f(x) = \sqrt {x - 7} $ on the interval $\left[ {11,23} \right]$
Here $a = 11{\text{ and b = 23}}$
Applying Mean Value Theorem
We can apply the Mean Value Theorem in the given expression two hypothesis such that
$f(a) = f(11) = \sqrt {11 - 7} $
Let us subtract we get, $ = \sqrt {16} $
$ = \sqrt 4 $
On splitting the term and we get,
$ = \sqrt {2 \times 2} $
Thus we get,
$ = 2$
$f(b) = f(23) = \sqrt {23 - 7} $
Let us subtract the term and we get,
$ = \sqrt {16} $
On splitting the term and we get,
$ = \sqrt {4 \times 4} $
Thus we get,
$ = 4$
This question is continuous on its domain, $[0,\infty )$, So it is continuous on $[11,23].$
Now we have to differentiate the given function as $f'(x) = \dfrac{1}{{2\sqrt {x - 7} }}$which exists for all $x > 7$(for all $x \in (7,\infty )$
So $f$ is differentiable on$(11,23)$
Now there must be exist at least one real number $c \in (11,23)$
$ \Rightarrow f'(c) = \dfrac{{f(23) - f(11)}}{{23 - 11}}$
Putting the values and we get,
$ = \dfrac{{4 - 2}}{{12}}$
On subtracting the numerator term and we get,
$ = \dfrac{2}{{12}}$
Canceling the numerator and denominator by $2$, we get
$ = \dfrac{1}{6}$
Again we have $f'(c) = \dfrac{1}{{2\sqrt {c - 7} }}$ Put $x = c$ in $f'(x)$we get
$f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$
Substituting the values for each expression
$\dfrac{1}{{2\sqrt {c - 7} }} = \dfrac{1}{6}$
Taking cross multiply we get,
$6 = 2\sqrt {c - 7} $
Squaring both sides, we get
$36 = 4(c - 7)$
Let us divide 4 on both sides and we get
$\dfrac{{36}}{4} = c - 7$
On dividing, we get
$9 = c - 7$
Taking one side the whole numbers
$9 + 7 = c$
By adding, we get
$16 = c$
$\therefore c = 16$ Which lies in the interval $[11,23]$
Hence the given function satisfies Mean value Theorem.

Note: Mean Value Theorem (MVT) is also known as LMVT (Lagrange’s Mean Value Theorem) which states that that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.
So to avoid mistakes one must retain that there is only one formula for LMVT and you have to use it properly. A student should differentiate and put the values carefully too.