How do you determine all values of c that satisfy the mean value theorem on the interval [1, 4] for #f(x)=1/sqrt(x)#?

Question

Cameron Alexander · Accepted Answer

undefined To determine all values of $ c $ that satisfy the Mean Value Theorem (MVT) on the interval $[1, 4]$ for $ f(x) = \frac{1}{\sqrt{x}} $, you need to find the average rate of change of the function over that interval and then find where this average rate of change equals the instantaneous rate of change.

The average rate of change of $ f(x) $ over the interval $[a, b]$ is given by:

$$ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} $$

For $ f(x) = \frac{1}{\sqrt{x}} $ on the interval $[1, 4]$, we have:

$$ f(1) = \frac{1}{\sqrt{1}} = 1 $$
$$ f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} $$

$$ \text{Average rate of change} = \frac{\frac{1}{2} - 1}{4 - 1} = -\frac{1}{6} $$

To apply the Mean Value Theorem, find the derivative of $ f(x) $ and set it equal to the average rate of change:

$$ f'(x) = \frac{-1}{2x^{3/2}} $$

Setting this equal to $ -\frac{1}{6} $ and solving for $ x $ gives:

$$ \frac{-1}{2x^{3/2}} = -\frac{1}{6} $$
$$ x^{3/2} = 3 $$
$$ x = 3^{\frac{2}{3}} $$

Thus, the value of $ c $ that satisfies the Mean Value Theorem on the interval $[1, 4]$ for $ f(x) = \frac{1}{\sqrt{x}} $ is $ c = 3^{\frac{2}{3}} $.

Cameron Alexander · Answer

The value of #c=2.08#

The mean value theorem states that if a function #f(x)# is continuous on the interval #[a,b]# and differentiable on the interval #(a,b)#, then

#EE c in [a,b]# such that

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

Here,

#f(x)=1/sqrtx#

The domain of #f(x)# is #D_f(x) in (0, +oo)#

The interval #[1,4] in D_f(x)#

#f'(x)=-x^(-3/2)/2#

#f'(c)=-c^(-3/2)/2#

#f(4)=1/sqrt4=1/2#

#f(1)=1/sqrt1=1#

Therefore,

#-c^(-3/2)/2=(1/2-1)/(4-1)=-1/6#

#c^(-3/2)=1/3#

#c^(3/2)=3#

#c^3=9#

#c=root(3)9=2.08#

#2.08 in [1,4]#