f(x)=⎩⎨⎧−x24−5x3x−4x2−12x+2xwhen x≤0when 0<x≤1when 1<x≤2when x>2
x→0−limf(x)=x→0−lim(−x2)=0
x→0+limf(x)=x→0+lim(4−5x)=4
x→0−limf(x)=0=4=x→0+limf(x) x→0limf(x) does not exist.
The function f(x) is discontinuous at x=0 and has a jump discontinuity at x=0.
x→1−limf(x)=x→1−lim(4−5x)=−1
x→1+limf(x)=x→1+lim(3x−4x2)=−1
x→1−limf(x)=−1=x→1+limf(x) x→1limf(x)=−1.
f(1)=4−5(1)=−1 The function f(x) is continuous at x=1.
x→2−limf(x)=x→2−lim(3x−4x2)=−10
x→2+limf(x)=x→2+lim(−12x+2x)=−20
x→2−limf(x)=−10=−20=x→2+limf(x) x→2limf(x) does not exist.
The function f(x) is discontinuous at x=2 and has a jump discontinuity at x=2.
f(1)=−1
h→0−limhf(1+h)−f(1)=h→0−limh4−5(1+h)−(−1)
=h→0−limh−5h=−5
h→0+limhf(1+h)−f(1)
=h→0+limh3(1+h)−4(1+h)2−(−1)
=h→0+limh3+3h−4−8h−4h2+1
=h→0+lim(−5−4h)=−5
h→0−limhf(1+h)−f(1)=−5
=h→0+limhf(1+h)−f(1) Then
h→0limhf(1+h)−f(1)=−5
The function f(x) is derivable at x=1, and
f′(1)=−5
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