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

Question

Ezekiel Alexander · Accepted Answer

The Mean Value Theorem has two hypotheses:

H1 : #f# is continuous on the closed interval #[a,b]#

H2 : #f# is differentiable on the open interval #(a,b)#.

#f# satisfies the hypotheses on #[a,b]# if and only if the hypotheses are true for #f# on #[a,b]# (and on #(a,b)# for H2.)

So you need to determine whether #f(x) = 6cosx# is continuous on #[-pi/2,pi/2]# and differentiable on #(-pi/2,pi/2)#.

(It is)

The conclusion on the Mean Value Theorem is

C : there is a #c# in #(a,b)# such that #f'(c) = (f(b)-f(a))/(b-a)#

To find the #c# mentioned in the conclusion, you need to solve the equation. Discard any solutions outside of #(a,b) = (-pi/2,pi/2)#.

(#c=0#)

Adam Adkins · Answer

undefined To determine if the function $ f(x) = 6 \cos(x) $ satisfies the hypotheses of the Mean Value Theorem on the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$, you need to check two conditions:

1. **Continuity:** Verify if $ f(x) $ is continuous on the given interval.
2. **Differentiability:** Confirm if $ f(x) $ is differentiable on the given interval.

For $ f(x) = 6 \cos(x) $ on the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$:

1. **Continuity:** The cosine function is continuous everywhere, so $ f(x) = 6 \cos(x) $ is continuous on the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$.

2. **Differentiability:** The cosine function is differentiable everywhere, so $ f(x) = 6 \cos(x) $ is differentiable on the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$.

Since both conditions are satisfied, the Mean Value Theorem applies.

To find the value of $ c $ guaranteed by the Mean Value Theorem, calculate the average rate of change of $ f(x) $ over the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$ using the formula:

$$ f' (c) = \frac{f(\frac{\pi}{2}) - f(-\frac{\pi}{2})}{\frac{\pi}{2} - (-\frac{\pi}{2})} $$

Evaluate $ f(\frac{\pi}{2}) $ and $ f(-\frac{\pi}{2}) $, then find the derivative of $ f(x) $, $ f'(x) $, and solve for $ c $.

William Abdalla · Answer

undefined To determine whether the function $ f(x) = 6 \cos(x) $ satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$ and find the value of $ c $, follow these steps:

1. **Continuity**:
   - The function $ f(x) = 6 \cos(x) $ is a composition of continuous functions (cosine is continuous everywhere), and thus $ f(x) $ is continuous on $[- \frac{\pi}{2}, \frac{\pi}{2}]$.

2. **Differentiability**:
   - To check for differentiability, compute the derivative $ f'(x) $ of $ f(x) $.
   $$ f'(x) = \frac{d}{dx} [6 \cos(x)] $$
   Using the chain rule:
   $$ f'(x) = -6 \sin(x) $$
   - $ f'(x) $ is continuous and exists everywhere in the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$.

3. **Conclusion**:
   - Since $ f(x) $ is continuous on $[- \frac{\pi}{2}, \frac{\pi}{2}]$ and $ f'(x) $ exists and is continuous on $[- \frac{\pi}{2}, \frac{\pi}{2}]$, $ f(x) $ satisfies the hypotheses of the Mean Value Theorem on this interval.

4. **Finding $ c $**:
   - According to the Mean Value Theorem, there exists at least one $ c $ in $[- \frac{\pi}{2}, \frac{\pi}{2}]$ such that:
   $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$
   where $ a = -\frac{\pi}{2} $ and $ b = \frac{\pi}{2} $.
   - First, compute $ f(b) $ and $ f(a) $:
   $$ f\left(\frac{\pi}{2}\right) = 6 \cos\left(\frac{\pi}{2}\right) = 0 $$
   $$ f\left(-\frac{\pi}{2}\right) = 6 \cos\left(-\frac{\pi}{2}\right) = 0 $$
   - Now, find the average rate of change:
   $$ \frac{f(b) - f(a)}{b - a} = \frac{0 - 0}{\frac{\pi}{2} + \frac{\pi}{2}} = 0 $$
   - To find $ c $, set $ f'(c) = 0 $:
   $$ -6 \sin(c) = 0 $$
   - The sine function is zero at $ c = 0 $ (among other values, but this is the only one in the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$).
   
5. **Conclusion**:
   - The function $ f(x) = 6 \cos(x) $ satisfies the hypotheses of the Mean Value Theorem on $[- \frac{\pi}{2}, \frac{\pi}{2}]$.
   - The value of $ c $ where the tangent is parallel to the secant line is $ c = 0 $.