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01 Definition of the derivative

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School
CUNY College of Staten Island**We aren't endorsed by this school
Course
MTH 232
Subject
Mathematics
Date
Dec 17, 2024
Pages
9
Uploaded by bingmo92
Topic: Definition of the derivativeQuestion: Use the definition of the derivative to find the simplified form of the limit.f(x) =x3-2xAnswer choices:A f(x) = limx0(x+x)3-2(x+x)xB f(x) = limx0(3x2+ 3xx+x2-2)C f(x) = limx0(x2-2)D f(x) = limx0(x+x)3-(x3-2x)x143
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Solution: BAfter replacing xwith (x+x)in f(x),f(x) = (x+x)3-2(x+x)we’ll substitute for f(x+x).f(x) = limx0(x+x)3-2(x+x)-f(x)xThen plug f(x)into the definition.f(x) = limx0(x+x)3-2(x+x)-(x3-2x)xf(x) = limx0x3+x2x+ 2x2x+ 2xx2+xx2+x3-2x-2x-x3+ 2xxCollect like terms,f(x) = limx0x2x+ 2x2x+ 2xx2+xx2+x3-2x-2x+ 2xxf(x) = limx0x2x+ 2x2x+ 2xx2+xx2+x3-2xxf(x) = limx03x2x+ 3xx2+x3-2xxthen factor xout of the numerator and cancel out that common factor from the numerator and denominator.f(x) = limx0x(3x2+ 3xx+x2-2)x144
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f(x) = limx0(3x2+ 3xx+x2-2)145
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Topic: Definition of the derivativeQuestion: Use the definition of the derivative to find the derivative of the function.f(x) =x2Answer choices:A f(x) = 0B f(x) = 2C f(x) = 2xD f(x) =x2+ 2x146
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Solution: CAfter replacing xwith (x+x)in f(x),f(x) = (x+x)2we’ll substitute for f(x+x).f(x) = limx0(x+x)2-f(x)xThen plug f(x)into the definition.f(x) = limx0(x+x)2-x2xf(x) = limx0x2+xx+xx+x2-x2xCollect like terms,f(x) = limx0xx+xx+x2xf(x) = limx02xx+x2xthen factor xout of the numerator and cancel out that common factor from the numerator and denominator.f(x) = limx0x(2x+x)xf(x) = limx0(2x+x)147
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Now we evaluate the limit using substitution, which means we’ll substitute x= 0.f(x) = 2x+ 0f(x) = 2x148
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Topic: Definition of the derivativeQuestion: Use the definition of the derivative to find the derivative of the function.f(x) = 2-x2+xAnswer choices:A f(x) = 2B f(x) = 2xC f(x) =-2xD f(x) =-2x+ 1149
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Solution: DAfter replacing xwith (x+x)in f(x),f(x) = 2-(x+x)2+ (x+x)we’ll substitute for f(x+x).f(x) = limx02-(x+x)2+ (x+x)-f(x)xThen plug f(x)into the definition.f(x) = limx02-(x+x)2+ (x+x)-(2-x2+x)xf(x) = limx02-(x2+xx+xx+x2) +x+x-2 +x2-xxf(x) = limx02-x2-xx-xx-x2+x+x-2 +x2-xxCollect like terms,f(x) = limx02-xx-xx-x2+x+x-2-xxf(x) = limx0-xx-xx-x2+x+x-xxf(x) = limx0-xx-xx-x2+xxf(x) = limx0-2xx-x2+xx150
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then factor xout of the numerator and cancel out that common factor from the numerator and denominator.f(x) = limx0x(-2x-x+ 1)xf(x) = limx0(-2x-x+ 1)Now we evaluate the limit using substitution, which means we’ll substitute x= 0.f(x) =-2x-0 + 1f(x) =-2x+ 1151
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