# `f(x) = cos(x) + tan(x), [0,pi]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c`...

`f(x) = cos(x) + tan(x), [0,pi]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the open interval `(a,b)` such that `f'(c) = (f(b) - f(a))/(b - a)`. If the Mean Value Theorem cannot be applied, explain why not.

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### Textbook Question

Chapter 3, 3.2 - Problem 46 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=cosx+tanx`

Mean value theorem can be applied,

1. if f is continuous on the closed interval `[a,b]` ,

2. if f is differentiable on the open interval (a,b)

3. there is a number c in (a,b) such that

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

Let us check the continuity of the function on the closed interval `[0,pi]` by plotting the graph of the function.

So the function is not continuous at x=`pi` /2

`f(pi/2)=cos(pi/2)+tan(pi/2)`

f(pi/2) is undefined

`f'(x)=-sin(x)+sec^2(x)`

So the function is differentiable.

`f'(pi)=cos(pi)+tan(pi)=-1`

`f'(0)=cos(0)+tan(0)=1`

`f'(c)=(f'(pi)-f(0))/(pi-0)`

`f'(c)=-2/pi`

Now there is no value of c in the closed interval `[0,pi]`  for which f'(c) is negative.

So the function does not satisfies the condition number 1 and 3. Hence Mean value theorem can not be applied.

Graph is attached Blue colour is for function and red for f'(x)

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