Given: f(x)= {1/x, if 0<x<b {1-1/(4x), if b less than or equal to x A. Determine a value of b so that f is continuous at b B. Is f differentiable at the value of b? Verify this using the definiton of derivative
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Given `f(x)= {[1/x,0<x<b],[1-1/(4x),b<=x]}.`
(a) `1/x` is continuous for all values of x, x>0. Also `1-1/(4x)` is continuous for all values of x>0.
For f(x) to be continuous at b, we need `1/x` evaluated at b to be equal to `1-1/(4x)` evaluated at b.
`1/x=1-1/(4x) ==> 4=4x-1 ==> x=5/4`
f(x) is continuous at b if `b=5/4`
(b) To determine if f(x) is differentiable at `b=5/4` :
`lim_(x->(5/4)^-)(f(x)-f(5/4))/(x-5/4)`
`=lim_(x->(5/4)^-)(1/x-4/5)/(x-5/4)`
`=lim_(x->(5/4)^-)((-4(x-5/4))/(5x))/(x-5/4)`
`=lim_(x->(5/4)^-)-4/(5x)=-16/25`
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`lim_(x->(5/4)^+)(f(x)-f(5/4))/(x-5/4)`
`=lim_(x->(5/4)^+)((1-1/(4x))-(1-1/(4(5/4))))/(x-5/4)`
`=lim_(x->(5/4)^+)(1-1/(4x)-4/5)/(x-5/4)`
`=lim_(x->(5/4)^+)((4x-5)/(20x))/(x-5/4)`
`=lim_(x->(5/4)^+)((4(x-5/4))/(20x))/(x-5/4)`
`=lim_(x->(5/4)^+)4/(20x)=4/25`
Since the left-hand limit does not agree with the right-hand limit, the function is not differentiable at `x=5/4`
The graph:
Note the "cusp" at `x=5/4`
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