Given the function #f(x)=(x-4)^2-1#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [3,0] and find the c?

Question

Emma Aldridge · Accepted Answer

#c=3/2# Mean value theorem states that for a function defined and continuous on #[a,b]# and continuously differentiable on #(a,b)# then there exists some #a < c < b# such that #f'(c) = (f(b)-f(a))/(b-a)# With #a = 0 and b = 3# we have #(((3-4)^2-1) - ((0-4)^2-1))/3 = -5# #f'(x) = 2(x-4)# #implies f'(c) = 2(c-4)# #2c - 8 = -5# #2c = 3 implies c = 3/2#

Christopher Allers · Answer

undefined To determine whether $ f(x) = (x - 4)^2 - 1 $ satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval $[3, 4]$, you need to check if the function is continuous on the closed interval and differentiable on the open interval.

1. **Continuity**: $ f(x) $ is a polynomial function, and polynomial functions are continuous everywhere. Therefore, $ f(x) $ is continuous on the closed interval $[3, 4]$.

2. **Differentiability**: To check differentiability, compute $ f'(x) $. Then check if $ f'(x) $ is continuous on the open interval $(3, 4)$.

$$ f(x) = (x - 4)^2 - 1 $$

$$ f'(x) = 2(x - 4) $$

$$ f'(x) = 2x - 8 $$

$ f'(x) $ is a linear function and is continuous everywhere. So, it is continuous on the open interval $(3, 4)$.

Since $ f(x) $ is both continuous on $[3, 4]$ and differentiable on $(3, 4)$, it satisfies the hypotheses of the Mean Value Theorem on the interval $[3, 4]$.

To find the value of $ c $ that satisfies the conclusion of the Mean Value Theorem, use the formula:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

Where $ a = 3 $ and $ b = 4 $.

$$ f'(c) = \frac{f(4) - f(3)}{4 - 3} $$

$$ f'(c) = \frac{(4 - 4)^2 - 1 - ((3 - 4)^2 - 1)}{1} $$

$$ f'(c) = \frac{-1 - (-4)}{1} $$

$$ f'(c) = \frac{3}{1} $$

$$ f'(c) = 3 $$

Now, solve for $ c $:

$$ 2c - 8 = 3 $$

$$ 2c = 11 $$

$$ c = \frac{11}{2} $$

So, $ c = \frac{11}{2} $ satisfies the conclusion of the Mean Value Theorem on the interval $[3, 4]$.