How do you determine all values of c that satisfy the mean value theorem on the interval [0,3] for #f(x) = x^3 + x - 1 #?

Question

Elijah Abram · Accepted Answer

#sqrt 3# Mean Value Theorem reads #f(3) = f(0) + 3 f'(c), cin (0, 3)# #f(x)=x^3+x-1 and f'(x)=3x^2+1#. #f(3) = 29, f(0)=-1 and f'(c)=3c^2+1#. So, #29 = -1 + 3(3c^2+1)# #c^2=3 and c=+-sqrt 3. -sqrt 3# is inadmissible.

Isaiah Allen · Answer

undefined To determine all values of $ c $ that satisfy the Mean Value Theorem on the interval $[0,3]$ for $ f(x) = x^3 + x - 1 $, we need to find the value of $ c $ where the instantaneous rate of change (derivative) of the function equals the average rate of change of the function over the interval $[0,3]$.

First, we find the derivative of $ f(x) $ with respect to $ x $:
$$ f'(x) = 3x^2 + 1 $$

Next, we find the average rate of change of $ f(x) $ over the interval $[0,3]$:
$$ \text{Average Rate of Change} = \frac{f(3) - f(0)}{3 - 0} = \frac{(3^3 + 3 - 1) - (0^3 + 0 - 1)}{3} = \frac{35}{3} $$

Then, we set the derivative equal to the average rate of change and solve for $ c $:
$$ 3c^2 + 1 = \frac{35}{3} $$
$$ 3c^2 = \frac{35}{3} - 1 $$
$$ 3c^2 = \frac{32}{3} $$
$$ c^2 = \frac{32}{9} $$
$$ c = \pm \sqrt{\frac{32}{9}} $$
$$ c = \pm \frac{4\sqrt{2}}{3} $$

Therefore, the values of $ c $ that satisfy the Mean Value Theorem on the interval $[0,3]$ for $ f(x) = x^3 + x - 1 $ are $ c = \frac{4\sqrt{2}}{3} $ and $ c = -\frac{4\sqrt{2}}{3} $.