Need help and want to know steps, plus which of the tattoo formulas i have to use if needed *TATTOO"CHEAT SHEET - USE OFTEN!youRTHE BASICSderv.PX) MPCO) MCRX) MRX→ 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})DERIVATIVESPRODUCTS AND QUOTIENTSu isAFUNCTION,X IS VAR.,e AND n y= u y'= u'v -uv'y=x^y'= nxn-iU AND Vyouv y'- u'v+uv'%3Dn-Iy'= nu" (u')y=e" y'- u'e"y=Lnu y'= 4 doutdiny=u^AREFUNCTIONSCONSTANTSLOGS AND EXPONENTSy=a" y'=a"u'ına ) WHERE Uy-a* y = a*x' In a fa is const,Fa*(1) inaINTEGRALSIS A FUNC,x^dx+K, n+-IX IS VARVABLEntldx → U'+k, n+-1Un(e*) =x (SIMPLIFIED, NOT DERIVATI VE )=X (SIMPUFIED, NOT DERIVATIVE)y= Ju'e"dr → e" + kin(MN) = Ln M+ inNin (A) - LnM -UnNin (M) = PlnMy= logax ay =xWHERE M» In u +K n=-lEN ARESTEPSFUNCTIONS.(CONVERSIONSNOT DERIVATIVĖS1) MAKE IT PRETTY. WHICH INTEGRAL?2) FIND U; CREATE U'3) WE HAVE4) MAKE IT LOOK LIKE TEMPLATE5) PERFORM INTEGRALWE WANT.CHANGE OF BASE:y=loga xlogaDEFINITE INTEGRALSlog XIn X%3DALSOUnay=Jax^dx = afx°dxScan"y= J(ax^+bx") dxSFondh Fo = Fb) - Fla)Jfandk=Fx)%3D%3Da 7. The magic (was it magic, really?) of the Black Friday Sale is over. Everyone sleeps in now, or shopsonline. Covid has taken its toll... No one breaks down the door at 4 am to get to the doorbuster deals. Butthere are still deals to be had, and there is a point in the day when the rate of sales is at its maximum. Ifthe store opens at 8 am, S represents sales after the doors open, and t represents every hour after 8 amwhen the doors open, find the time of day during its morning sales when the rate of sales is at itsmaximum. In other words, find the point of diminishing returns. Make sure to graph the originalfunction, with 0

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

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

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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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.

Question

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

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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