Question #185363

determine the point of discontinuity of the function f and the nature of discontinuity at each points

f(x)= { -x^2, when x<=0

{ 4-5x, when 0<x<=1

{ 3x-4x^2 when 1<x<=2

{-12x+2x when x>2

and check whether function f is derivable at x=1


1
Expert's answer
2021-05-07T09:02:22-0400
f(x)={x2when x045xwhen 0<x13x4x2when 1<x212x+2xwhen x>2f(x) = \begin{cases} -x^2 &\text{when } x\leq0 \\ 4-5x &\text{when } 0<x\leq1\\ 3x-4x^2 &\text{when } 1<x\leq2\\ -12x+2x &\text{when } x>2\\ \end{cases}


limx0f(x)=limx0(x2)=0\lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}(-x^2)=0

limx0+f(x)=limx0+(45x)=4\lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}(4-5x)=4

limx0f(x)=04=limx0+f(x)\lim\limits_{x\to0^-}f(x)=0\not=4=\lim\limits_{x\to0^+}f(x)

limx0f(x)\lim\limits_{x\to0}f(x) does not exist.

The function f(x)f(x) is discontinuous at x=0x=0 and has a jump discontinuity at x=0.x=0.



limx1f(x)=limx1(45x)=1\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}(4-5x)=-1

limx1+f(x)=limx1+(3x4x2)=1\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}(3x-4x^2)=-1

limx1f(x)=1=limx1+f(x)\lim\limits_{x\to1^-}f(x)=-1=\lim\limits_{x\to1^+}f(x)

limx1f(x)=1\lim\limits_{x\to1}f(x)=-1.


f(1)=45(1)=1f(1)=4-5(1)=-1

The function f(x)f(x) is continuous at x=1.x=1.



limx2f(x)=limx2(3x4x2)=10\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(3x-4x^2)=-10

limx2+f(x)=limx2+(12x+2x)=20\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(-12x+2x)=-20

limx2f(x)=1020=limx2+f(x)\lim\limits_{x\to2^-}f(x)=-10\not=-20=\lim\limits_{x\to2^+}f(x)

limx2f(x)\lim\limits_{x\to2}f(x) does not exist.

The function f(x)f(x) is discontinuous at x=2x=2 and has a jump discontinuity at x=2.x=2.



f(1)=1f(1)=-1


limh0f(1+h)f(1)h=limh045(1+h)(1)h\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0^-}\dfrac{4-5(1+h)-(-1)}{h}

=limh05hh=5=\lim\limits_{h\to0^-}\dfrac{-5h}{h}=-5

limh0+f(1+h)f(1)h\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}

=limh0+3(1+h)4(1+h)2(1)h=\lim\limits_{h\to0^+}\dfrac{3(1+h)-4(1+h)^2-(-1)}{h}

=limh0+3+3h48h4h2+1h=\lim\limits_{h\to0^+}\dfrac{3+3h-4-8h-4h^2+1}{h}

=limh0+(54h)=5=\lim\limits_{h\to0^+}(-5-4h)=-5


limh0f(1+h)f(1)h=5\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=-5

=limh0+f(1+h)f(1)h=\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}

Then


limh0f(1+h)f(1)h=5\lim\limits_{h\to0}\dfrac{f(1+h)-f(1)}{h}=-5

The function f(x)f(x) is derivable at x=1,x=1, and


f(1)=5f'(1)=-5


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