According to the definition, the left hand and right hand derivatives of f at a are defined by
f′−(a)=limh→0−f(a+h)−f(a)h and f′+(a)=limh→0+f(a+h)−f(a)h
Suppose that these limits exists.
a.) Determine f′−(4) and f′+(4) for the function
f(x)={0ifx≤05−xif0<x<415−xifx≥4
b.) Sketch the graph of f
c.) Where is f discontinuous?
d.) Where is f not differentiable?
a.)
For Left Hand,
f′−(x)=limh→0−[5−(x+h)−(5−x)h]f′−(x)=limh→0−[\cancel5−\cancelx−h−\cancel5+\cancelxh]f′−(x)=limh→0−−\cancelh\cancelhf′−(x)=limh→0−−1f′−(4)=−1
For Right Hand,
f′+(a)=limh→0+[15−(x+h)−(15−x)]hf′+(a)=limh→0+5−x−[5−(x+h)](5−(x+h))(5−x)hf′+(a)=limh→0+\cancel5−\cancelx−\cancel5+\cancelx+h(5−(x+h))(5−x)hf′+(a)=limh→0+\cancelh(5−x−h)(5−x)\cancelhf′+(a)=limh→0+1(5−x−h)(5−x)f′+(a)=limh→0+1(5−x−0)(5−x)f′+(a)=limh→0+1(5−x)2f′+(4)=1(5−4)2=1(1)12=1f′+(4)=1
Therefore, f′(4) doesn't exist since f′−(4)≠f′+(4)
b.)
c.) Based from the graph
The function f is discontinuous at x=0 because of jump discontinuity.
Also, f is discontinuous at x=5 because of infinite discontinuity
d.) The function f is not differentiable at x=0 and x=5 because the function is discontinuous
at that points. Also, the function is not differentiable at x=4 since f′−(4)≠f′+(4) based
from definition.
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