Question

In need of some serious calculus help

1.) Determine whether Rolle's Theorem can be applied to f(x)=3-|x-3| on the interval [0,6]. If so find all values of x in (0,6) such that f'(c)=0. If not explain why not.

2.) Determine whether the Mean Value Theorem can be applied to f(x)=sqrt(2-x) on [-7,2]. If so find all values of c in (-7,2) such that f'(c)=[f(b)-f(a)]/(b-a). If not explain why not.

Answer 1

\[\frac{ d }{ dx }\left| f \left( x \right) \right|=\frac{ f \left( x \right) }{ \left| f \left( x \right) \right| }f \prime \left( x \right)\]
derivative does not exist at x=3, hence rolle's theorem cannot be applied on [0,6]

Answer 2

2.\[f \prime \left( c \right)=\frac{ f \left( b \right)-f \left( a \right) }{ b-a }\]
\[f \prime \left( x \right)=\frac{ -1 }{ 2\sqrt{2-x} },f \prime \left( c \right)=\frac{ -1 }{ 2\sqrt{2-c} }\]
\[-\frac{ 1 }{ 2\sqrt{2-c} }=\frac{ \sqrt{2-2} -\sqrt{2-\left( -7 \right)}}{ 2-\left( -7 \right) }\]
\[-\frac{ 1 }{ 2\sqrt{2-c} }=\frac{ -3 }{ 9 }=-\frac{ 1 }{ 3 }\]
\[2\sqrt{2-c}=3,2-c=\frac{ 9 }{ 4 },c=2-\frac{ 9 }{ 4 }=-\frac{ 1 }{ 4 }\in \left( -7,2 \right)\]
hence M.V.T. is verified.
when \[x=-\frac{ 1 }{ 4 },f \left( x \right)=\sqrt{2-\left( -\frac{ 1 }{ 4 } \right)}=\frac{ 3 }{ 2 }\]
Hence tangent at \[c\left( -\frac{ 1 }{ 4 },\frac{ 3 }{ 2 } \right)~is~\parallel~\to~the~chord~joining~\left[ -7,2 \right]\]

Answer 3

Thank you!!

Answer 4

yw