Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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Need help and want to know steps, plus which of the tattoo formulas i have to use if needed
*TATTOO"CHEAT SHEET - USE OFTEN! youR THE BASICS derv. PX) MP CO) MC RX) MR X→ fcx)+y (ly CAN BE +,Ø,-) X f'x) SLOPE(SLOPE CAN BE +,Ø, integ. - {0 = MAx OR MIN OR H.P.I.3) X *f"(x) + CONCAVITY (CONCAVITY IS ,A,Ø { Bis POINT OF INFLECTION}) DERIVATIVES PRODUCTS AND QUOTIENTS u isA FUNCTION, X IS VAR., e AND n y= u y'= u'v -uv' y=x^ y'= nxn-i U AND V youv y'- u'v+uv' %3D n-I y'= nu" (u') y=e" y'- u'e" y=Lnu y'= 4 doutdin y=u^ ARE FUNCTIONS CONSTANTS LOGS AND EXPONENTS y=a" y'=a"u'ına ) WHERE U y-a* y = a*x' In a fa is const, Fa*(1) ina INTEGRALS IS A FUNC, x^dx +K, n+-I X IS VARVABLE ntl dx → U' +k, n+-1 Un(e*) =x (SIMPLIFIED, NOT DERIVATI VE ) =X (SIMPUFIED, NOT DERIVATIVE) y= Ju'e"dr → e" + k in(MN) = Ln M+ inN in (A) - LnM -UnN in (M) = PlnM y= logax ay =x WHERE M » In u +K n=-l EN ARE STEPS FUNCTIONS. (CONVERSIONS NOT DERIVATIVĖS 1) MAKE IT PRETTY. WHICH INTEGRAL? 2) FIND U; CREATE U' 3) WE HAVE 4) MAKE IT LOOK LIKE TEMPLATE 5) PERFORM INTEGRAL WE WANT. CHANGE OF BASE: y=loga x loga DEFINITE INTEGRALS log X In X %3D ALSO Una y=Jax^dx = afx°dx Scan" y= J(ax^+bx") dx SFondh Fo = Fb) - Fla) Jfandk=Fx) %3D %3D a
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Transcribed Image Text:*TATTOO"CHEAT SHEET - USE OFTEN! youR THE BASICS derv. PX) MP CO) MC RX) MR X→ fcx)+y (ly CAN BE +,Ø,-) X f'x) SLOPE(SLOPE CAN BE +,Ø, integ. - {0 = MAx OR MIN OR H.P.I.3) X *f"(x) + CONCAVITY (CONCAVITY IS ,A,Ø { Bis POINT OF INFLECTION}) DERIVATIVES PRODUCTS AND QUOTIENTS u isA FUNCTION, X IS VAR., e AND n y= u y'= u'v -uv' y=x^ y'= nxn-i U AND V youv y'- u'v+uv' %3D n-I y'= nu" (u') y=e" y'- u'e" y=Lnu y'= 4 doutdin y=u^ ARE FUNCTIONS CONSTANTS LOGS AND EXPONENTS y=a" y'=a"u'ına ) WHERE U y-a* y = a*x' In a fa is const, Fa*(1) ina INTEGRALS IS A FUNC, x^dx +K, n+-I X IS VARVABLE ntl dx → U' +k, n+-1 Un(e*) =x (SIMPLIFIED, NOT DERIVATI VE ) =X (SIMPUFIED, NOT DERIVATIVE) y= Ju'e"dr → e" + k in(MN) = Ln M+ inN in (A) - LnM -UnN in (M) = PlnM y= logax ay =x WHERE M » In u +K n=-l EN ARE STEPS FUNCTIONS. (CONVERSIONS NOT DERIVATIVĖS 1) MAKE IT PRETTY. WHICH INTEGRAL? 2) FIND U; CREATE U' 3) WE HAVE 4) MAKE IT LOOK LIKE TEMPLATE 5) PERFORM INTEGRAL WE WANT. CHANGE OF BASE: y=loga x loga DEFINITE INTEGRALS log X In X %3D ALSO Una y=Jax^dx = afx°dx Scan" y= J(ax^+bx") dx SFondh Fo = Fb) - Fla) Jfandk=Fx) %3D %3D a
7. The magic (was it magic, really?) of the Black Friday Sale is over. Everyone sleeps in now, or shops online. Covid has taken its toll... No one breaks down the door at 4 am to get to the doorbuster deals. But there are still deals to be had, and there is a point in the day when the rate of sales is at its maximum. If the store opens at 8 am, S represents sales after the doors open, and t represents every hour after 8 am when the doors open, find the time of day during its morning sales when the rate of sales is at its maximum. In other words, find the point of diminishing returns. Make sure to graph the original function, with 0<t<4 hours. Mark the point of diminishing returns on the graph, and LABEL your axes with maxima and minima to show scale. S(t) = -4t3 + 24t2 + 50 What time of day is the rate of sales at its maximum?
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Transcribed Image Text:7. The magic (was it magic, really?) of the Black Friday Sale is over. Everyone sleeps in now, or shops online. Covid has taken its toll... No one breaks down the door at 4 am to get to the doorbuster deals. But there are still deals to be had, and there is a point in the day when the rate of sales is at its maximum. If the store opens at 8 am, S represents sales after the doors open, and t represents every hour after 8 am when the doors open, find the time of day during its morning sales when the rate of sales is at its maximum. In other words, find the point of diminishing returns. Make sure to graph the original function, with 0<t<4 hours. Mark the point of diminishing returns on the graph, and LABEL your axes with maxima and minima to show scale. S(t) = -4t3 + 24t2 + 50 What time of day is the rate of sales at its maximum?
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