Description
Online homework , quiz and final exam review quiz.
I need you to get a copy of septs for final's answers
About the homework, you are going to see the all homewrok, but most of them are done by me, there is a few of assignments
Explanation & Answer
Alright, here you go! :) I attached it as a PDF. I show the steps for every question. I hope it's helpful! I worked really hard on it haha.So I completed all the homework, the quiz, and the final exam review quiz and got 100% on all of them for you. I also wrote up the step-by-step solutions for the final exam review quiz like you asked. If you have any questions about it, let me know! It was great working with you. :)
Problem #1
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)
𝒍𝒊𝒎
𝒙→𝟐
𝒙𝟐 − 𝟔𝒙 + 𝟖
𝒙−𝟐
Factor and cancel.
𝑥 2 − 6𝑥 + 8 (𝑥 − 2)(𝑥 − 4)
=
=𝑥−4
𝑥−2
𝑥−2
𝑥 2 − 6𝑥 + 8
𝑙𝑖𝑚
= 𝑙𝑖𝑚 (𝑥 − 4) = 2 − 4 = −2
𝑥→2
𝑥→2
𝑥−2
Answer: −𝟐
Problem #2
For what value of the constant 𝒄 is the function 𝐟 continuous on (−∞, ∞)?
𝒇(𝒙) = {
𝒄𝒙𝟐 + 𝟔𝒙
𝒙𝟑 − 𝒄𝒙
𝒊𝒇 𝒙 < 𝟓
𝒊𝒇 𝒙 ≥ 𝟓
Set each part equal to each other, plug in 𝑥 = 5, and solve for 𝑐.
𝑐𝑥 2 + 6𝑥 = 𝑥 3 − 𝑐𝑥
𝑐(5)2 + 6(5) = (5)3 − 𝑐(5)
25𝑐 + 30 = 125 − 5𝑐
30𝑐 = 95
95 19
𝑐=
=
30
6
𝐀𝐧𝐬𝐰𝐞𝐫: 𝒄 =
𝟏𝟗
𝟔
Problem #3
Find the horizontal and vertical asymptotes of the curve.
𝒚=
𝟓𝒙𝟐 + 𝒙 − 𝟒
𝒙𝟐 + 𝒙 − 𝟑𝟎
First, factor to see if anything cancels.
𝑦=
5𝑥 2 + 𝑥 − 4 (5𝑥 + 4)(𝑥 − 1)
=
𝑥 2 + 𝑥 − 30
(𝑥 + 6)(𝑥 − 5)
Nothing cancels, so continue like normal.
Vertical asymptotes are when the denominator equals 0. So set denominator equal to 0 and
solve for 𝑥.
(𝑥 + 6)(𝑥 − 5) = 0
This happens when either 𝑥 + 6 = 0 or 𝑥 − 5 = 0. So we have two equations for our
vertical asymptotes:
𝑥 + 6 = 0 → 𝑥 = −6 and 𝑥 − 5 = 0 → 𝑥 = 5
To find horizontal asymptotes, we take the limits to −∞ and ∞.
5𝑥 2 + 𝑥 − 4
𝑙𝑖𝑚
𝑥→∞ 𝑥 2 + 𝑥 − 30
When taking a limit to −∞ or ∞ of a rational function, we only pay attention to the highest
degree terms and ignore the rest. So we can write this as:
𝑙𝑖𝑚
𝑥→∞
5𝑥 2 + 𝑥 − 4
5𝑥 2
=
𝑙𝑖𝑚
= 𝑙𝑖𝑚 (5) = 5
𝑥→∞
𝑥 2 + 𝑥 − 30 𝑥→∞ 𝑥 2
The limit to −∞ is the same.
5𝑥 2 + 𝑥 − 4
5𝑥 2
𝑙𝑖𝑚
= 𝑙𝑖𝑚
= 𝑙𝑖𝑚 (5) = 5
𝑥→−∞ 𝑥 2 + 𝑥 − 30
𝑥→−∞ 𝑥 2
𝑥→−∞
So our horizontal asymptote is 𝑦 = 5.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝒙 = −𝟔
𝒙=𝟓
𝒚=𝟓
Problem #4
Differentiate.
𝒚=
𝒙𝟑
𝒙+𝟒
+𝒙−𝟑
𝑓 ′ 𝑓 ′ 𝑔 − 𝑔′𝑓
Use the Quotient Rule: ( ) =
𝑔
𝑔2
*Tip: An easy way to remember this is “fig minus gif over gg”.
𝑓(𝑥) = 𝑥 + 4 → 𝑓 ′ (𝑥) = 1
𝑔(𝑥) = 𝑥 3 + 𝑥 − 3 → 𝑔′ (𝑥) = 3𝑥 2 + 1
𝑓 ′ 𝑓 ′ 𝑔 − 𝑔′𝑓 (1)(𝑥 3 + 𝑥 − 3) − (3𝑥 2 + 1)(𝑥 + 4)
( ) =
=
(𝑥 3 + 𝑥 − 3)2
𝑔
𝑔2
Multiply out the numerator and simplify.
𝑥 3 + 𝑥 − 3 − 3𝑥 3 − 12𝑥 2 − 𝑥 − 4 −2𝑥 3 − 12𝑥 2 − 7
=
(𝑥 3 + 𝑥 − 3)2
(𝑥 3 + 𝑥 − 3)2
𝐀𝐧𝐬𝐰𝐞𝐫:
−𝟐𝒙𝟑 − 𝟏𝟐𝒙𝟐 − 𝟕
(𝒙𝟑 + 𝒙 − 𝟑)𝟐
Problem #5
Differentiate. (Assume 𝒄 is a constant.)
𝒛 = 𝒗𝟑/𝟐 (𝒗 + 𝒄𝒆𝒗 )
Use the Product Rule: (𝑓𝑔)′ = 𝑓 ′ 𝑔 + 𝑔′𝑓
*Tip: An easy way to remember this is “fig plus gif”.
3
𝑓(𝑣) = 𝑣 3/2 → 𝑓 ′ (𝑣) = 𝑣 1/2
2
𝑔(𝑣) = 𝑣 + 𝑐𝑒 𝑣 → 𝑔′ (𝑣) = 1 + 𝑐𝑒 𝑣
3
(𝑓𝑔)′ = 𝑓 ′ 𝑔 + 𝑔′ 𝑓 = ( 𝑣 1/2 ) (𝑣 + 𝑐𝑒 𝑣 ) + ( 1 + 𝑐𝑒 𝑣 )( 𝑣 3/2 )
2
Multiply out.
3 3/2 3𝑐 1/2 𝑣
𝑣 + 𝑣 𝑒 + 𝑣 3/2 + 𝑐𝑣 3/2 𝑒 𝑣
2
2
This doesn’t simplify.
𝐀𝐧𝐬𝐰𝐞𝐫:
𝟑 𝟑/𝟐 𝟑𝒄 𝟏/𝟐 𝒗
𝒗 + 𝒗 𝒆 + 𝒗𝟑/𝟐 + 𝒄𝒗𝟑/𝟐 𝒆𝒗
𝟐
𝟐
Problem #6
Differentiate the following function.
𝒉(𝜽) = 𝟒 𝒔𝒆𝒄(𝜽) + 𝟓𝒆𝜽 𝒕𝒂𝒏(𝜽)
Use the Sum Rule: (𝑓 + 𝑔)′ = 𝑓 ′ + 𝑔′
𝑓(𝜃) = 4 sec(𝜃) → 𝑓 ′ (𝜃) = 4 sec(𝜃) tan(𝜃)
𝑔(𝜃) = 5𝑒 𝜃 tan(𝜃) → 𝑔′ (𝜃) = We have to use the Product Rule to find 𝑔′ (𝜃).
Product Rule (use different letters since we’re already using 𝑓 and 𝑔): (ℎ𝑗)′ = ℎ′ 𝑗 + 𝑗 ′ ℎ
ℎ(𝜃) = 5𝑒 𝜃 → ℎ′ (𝜃) = 5𝑒 𝜃
𝑗(𝜃) = tan(𝜃) → 𝑗 ′ (𝜃) = sec 2 (𝜃)
𝑔′ = (ℎ𝑗)′ = ℎ′ 𝑗 + 𝑗 ′ ℎ = (5𝑒 𝜃 )( tan(𝜃)) + (sec 2 (𝜃))(5𝑒 𝜃 ) = 5𝑒 𝜃 ( tan(𝜃) + sec 2 (𝜃))
So (𝑓 + 𝑔)′ = 𝑓 ′ + 𝑔′ = 4 sec(𝜃) tan(𝜃) + 5𝑒 𝜃 ( tan(𝜃) + sec 2 (𝜃))
We can plug this into a calculator and get:
4 sec(𝜃) tan(𝜃) + 5𝑒 𝜃 ( tan(𝜃) + sec 2 (𝜃)) =
5𝑒 𝜃 sin(𝜃) cos(𝜃) + 4 sin(𝜃) + 5𝑒 𝜃
, but that
cos2 (𝜃)
isn’t necessary, we can just leave it like it is.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒 𝐬𝐞𝐜(𝜽) 𝐭𝐚𝐧(𝜽) + 𝟓𝒆𝜽 ( 𝐭𝐚𝐧(𝜽) + 𝐬𝐞𝐜 𝟐 (𝜽))
Problem #7
Find 𝒚′ and 𝒚′′.
𝒚 = 𝒄𝒐𝒔(𝒙𝟐 )
′
Use the Chain Rule: (𝑓(𝑔(𝑥))) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′(𝑥)
*Tip: You can think of this as “the derivative of the outside function times the derivative of
the inside function”.
𝑓(𝑥) = cos(𝑥) → 𝑓 ′ (𝑥) = − sin(𝑥)
𝑔(𝑥) = 𝑥 2 → 𝑔′ (𝑥) = 2𝑥
′
So (𝑓(𝑔(𝑥))) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′ (𝑥) = − sin(𝑔(𝑥)) 𝑔′ (𝑥) = − sin(𝑥 2 ) ∙ (2𝑥) = −2𝑥 sin(𝑥 2 )
To find the derivative of this, we need to use the Product Rule: (𝑓𝑔)′ = 𝑓 ′ 𝑔 + 𝑔′𝑓
(Note: This is a new 𝑓 and 𝑔.)
𝑓(𝑥) = −2𝑥 → 𝑓 ′ (𝑥) = −2
𝑔(𝑥) = sin(𝑥 2 ) → 𝑔′ (𝑥) = We have to use the Chain Rule to find 𝑔′ (𝑥).
Chain Rule (use different letters since we’re already using 𝑓 and 𝑔):
′
𝑔′ (𝑥) = (ℎ(𝑗(𝑥))) = ℎ′ (𝑗(𝑥)) ∙ 𝑗 ′ (𝑥)
ℎ(𝑥) = sin(𝑥) → ℎ′ (𝑥) = cos(𝑥)
𝑗(𝑥) = 𝑥 2 → 𝑗 ′ (𝑥) = 2𝑥
′
𝑔′ (𝑥) = (ℎ(𝑗(𝑥))) = ℎ′ (𝑗(𝑥)) ∙ 𝑗 ′ (𝑥) = cos(𝑗(𝑥)) ∙ 𝑗 ′ (𝑥) = cos(𝑥 2 ) ∙ (2𝑥) = 2𝑥 cos(𝑥 2 )
So (𝑓𝑔)′ = 𝑓 ′ 𝑔 + 𝑔′ 𝑓 = (−2)(sin(𝑥 2 )) + 2𝑥 cos(𝑥 2 ))(−2𝑥) = −2 sin(𝑥 2 ) − 4𝑥 2 cos(𝑥 2 )
Answer: 𝒚′ = −𝟐𝒙 𝐬𝐢𝐧(𝒙𝟐 )
𝒚′′ = −𝟐 𝐬𝐢𝐧(𝒙𝟐 ) − 𝟒𝒙𝟐 𝐜𝐨𝐬(𝒙𝟐 )
Problem #8
Find 𝒅𝒚/𝒅𝒙 ...