Find value(s) c ∈ (−2, 4) such that f'(c) is parallel to the chord line joining ?

Question

Im not sure wich Theorem they mean :/

Emma Aldridge · Accepted Answer

Mean Value Theorem
If a function #f# is continuous on #[a,b]# and differentiable on #(a,b)#,
then there exists c in #(a,b)# such that #f'(c)={f(b)-f(a)}/{b-a}#.

You can learn more about the theorem here: https://tutor.hix.ai In this particular question, #f# is continuous on #[-2,4]#, but not differentiable on #(-2,4)# because the derivative does not exist at #1/2#. In this case there is no point in the interval at which the tangent line is parallel to the secant line (chord).

Scarlett Allison · Answer

undefined To find the value(s) of $ c $ such that $ f'(c) $ is parallel to the chord line joining two points $ (a, f(a)) $ and $ (b, f(b)) $, where $ a $ and $ b $ belong to the interval $ (-2, 4) $, you can follow these steps:

1. Find the slope of the chord line using the formula:
   $$ \text{Slope of chord} = \frac{f(b) - f(a)}{b - a} $$

2. Compute $ f'(x) $ using the derivative of the function $ f(x) $.

3. Set the derivative equal to the slope of the chord line:
   $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

4. Solve the equation obtained in step 3 to find the value(s) of $ c $ within the interval $ (-2, 4) $.

This value(s) of $ c $ will make $ f'(c) $ parallel to the chord line joining $ (a, f(a)) $ and $ (b, f(b)) $.