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Question: How do you determine if Rolle's theorem can be applied to \(f\left( x \right) = \sin 2x\) on the int...

How do you determine if Rolle's theorem can be applied to f(x)=sin⁡2xf\left( x \right) = \sin 2x on the interval [0,π2]\left[ {0,\dfrac{\pi }{2}} \right] and if so, how do you find all the values of cc in the interval for which f′(c)=0f'\left( c \right) = 0?

Explanation

Solution

In order to solve for Rolle's theorem, first we should know what Rolle's theorem is, then according to Rolle's theorem, prove the first three necessary rules. Equate the derivative of the given function with zero and get the value for cc in the given interval (a,b)\left( {a,b} \right).

Formula used:

  1. f′(sin⁡x)=cos⁡xf'\left( {\sin x} \right) = \cos x
  2. sin⁡0=0\sin 0 = 0
  3. sin⁡π=0\sin \pi = 0
  4. cos⁡π2=0\cos \dfrac{\pi }{2} = 0

Complete step by step solution:
We are given with the function: f(x)=sin⁡2xf\left( x \right) = \sin 2x.
Since, we know that, the Rolle's theorem says that if:

  1. y=f(x)y = f\left( x \right) is a continuous function in a set [a,b]\left[ {a,b} \right];
  2. y=f(x)y = f\left( x \right) is a derivable function in a set (a,b)\left( {a,b} \right);
  3. f(a)=f(b)f\left( a \right) = f\left( b \right); Then, there exists a c∈(a,b)c \in \left( {a,b} \right)as if f′(c)=0f'\left( c \right) = 0 exists.
    So, for solving our function with Rolle's theorem, we need to check all the three steps and are as follows:
  4. y=f(x)=sin⁡2xy = f\left( x \right) = \sin 2x, since we know that the function sine is continuous everywhere, so according to that f(x)=sin⁡2xf\left( x \right) = \sin 2x is also continuous in the interval [0,π2]\left[ {0,\dfrac{\pi }{2}} \right]. That implies the first point proved.
  5. Differentiating f(x)=sin⁡2xf\left( x \right) = \sin 2x and we get f′(x)=2cos⁡2xf'\left( x \right) = 2\cos 2x, which is also continuous everywhere, so our function is also differentiable everywhere. That gives our function is also differentiable at (0,π2)\left( {0,\dfrac{\pi }{2}} \right).
  6. Substituting 00 in f(x)=sin⁡2xf\left( x \right) = \sin 2x and we get: f(0)=sin⁡2(0)=0f\left( 0 \right) = \sin 2\left( 0 \right) = 0, Then substituting π2\dfrac{\pi }{2} in f(x)=sin⁡2xf\left( x \right) = \sin 2x and we get: f(π2)=sin⁡2.π2=sin⁡π=0f\left( {\dfrac{\pi }{2}} \right) = \sin 2.\dfrac{\pi }{2} = \sin \pi = 0. And we obtained that f(a)=f(b) ⇒f(0)=f(π2) f\left( a \right) = f\left( b \right) \\\ \Rightarrow f\left( 0 \right) = f\left( {\dfrac{\pi }{2}} \right) \\\
    And, this way our Rolle's theorem is verified, so there exists a c∈(a,b)c \in \left( {a,b} \right)as if f′(c)=0f'\left( c \right) = 0exists.
    Now, derivating f(x)=sin⁡2xf\left( x \right) = \sin 2x which can be written as f(c)=sin⁡2cf\left( c \right) = \sin 2c, with respect to cc, and we get:
    f′(c)=2cos⁡2cf'\left( c \right) = 2\cos 2c
    Comparing f′(c)f'\left( c \right) with zero, f′(c)=0f'\left( c \right) = 0:
    f′(c)=0 2cos⁡2c=0 cos⁡2c=0  f'\left( c \right) = 0 \\\ 2\cos 2c = 0 \\\ \cos 2c = 0 \\\
    Since, we know that cos⁡π2=0\cos \dfrac{\pi }{2} = 0, so substituting it in the above equation, we get:
    cos⁡2c=0 cos⁡2c=cos⁡π2 ⇒2c=π2 ⇒c=π4  \cos 2c = 0 \\\ \cos 2c = \cos \dfrac{\pi }{2} \\\ \Rightarrow 2c = \dfrac{\pi }{2} \\\ \Rightarrow c = \dfrac{\pi }{4} \\\
    Since, we can see that c=π4c = \dfrac{\pi }{4}lies in the interval c∈(0,π2)c \in \left( {0,\dfrac{\pi }{2}} \right).
    Therefore, Rolle's theorem can be applied to the function to f(x)=sin⁡2xf\left( x \right) = \sin 2x on the interval [0,π2]\left[ {0,\dfrac{\pi }{2}} \right] and value of cc for which f′(c)=0f'\left( c \right) = 0 is c=π4c = \dfrac{\pi }{4} in the interval (0,π2)\left( {0,\dfrac{\pi }{2}} \right).

Note:

  1. It’s important to check for the Rolle's theorem rules step by step, to solve this kind of function.
  2. The square bracket represents the closed interval whereas the round bracket represents the open intervals.