CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Verify Rolle's theorem\quad for the function $$f(x)={ e }^{ -x }\sin { x } ,x\epsilon \left[ 0,\pi  \right] $$


Solution

$$f\left( x \right) ={ e }^{ -x }\sin x\quad ,x\in (0,{ \pi  })$$
For Rolle's Theorem, $$f(0)=f(\pi)$$ & $$f(x)$$ must be continuous & differentiable over $$[0,\pi]$$
Let us check if $$f(0)=f(\pi)$$ 
$$\Rightarrow f(0)=\quad { e }^{ -0 }\sin (0)=\quad 0\\ \Rightarrow f(\pi )=\quad { e }^{ -\pi  }\sin (\pi )=\quad 0$$
Therefore, $$f(0)=f(\pi)$$ 
The function $${e}^{-x}$$ & $$\sin x$$ are both continuous & differentiable over $$[0,\pi]$$
Therefore, Rolle's Theorem can be applied for the function given.
There exists $$'c'$$ such that $$f'(c)=0$$
$$f'\left( x \right) =-{ e }^{ -x }\sin x+{ e }^{ -x }\cos x\\ f'\left( c \right) =-{ e }^{ -c }\sin (c)+{ e }^{ -c }\cos (c)=0\quad \Rightarrow { e }^{ -c }\left[ \cos c-\sin c \right] =0\\ \Rightarrow \cos c=\sin c\Rightarrow tanc=1\Rightarrow \boxed { c=\dfrac { \pi  }{ 4 }  } $$
Hence, Rolle's Theorem is verified.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image