I need help with these calculus questions
1. An equation of the line tangent to the graph of y=5x-1/3x+1 at the point where x = 1 is:
A) 7x+y=0
B) 9x+7y= 53
C) 2x-3y=13
D) x-2y=-1
2.For which value of x does the function f(x)=(x+2)(x+3)^2 have a relative maximum?
A)-2
B)-3
C)both -2 and -3
D)-5/2
3. How many critical points does the function f(x)=(x-1)^6 (x+5)^7 have?
A)13
B)7
C)6
D)3
1) Its pretty straight forward. To get slope of your tangent line find the derivative of your equation, which is in this case, 8 / (3x+1)^2. Put your x-value into this to find your slope at that point. Then plug value of x = 1 to your original equation to get your y-value for the tangent line equation. Afterwards use the formula y1 - y = m(x-x1) to get your equation using the values that you found for x and y.
m = 8/16
y = 1
x = 1
y1 -y = m(x-x1)
1 - y = 8/16(x-1)
y = -((8/16)x - 8/16 + 1)
y = -(1/2)x - 1/2
ANS: d
#1
take the derivative, sub in x = 1
Now you have the slope of the tangent
Sub x=1 into the original equation to get y
So now you have the slope and a point. Use the method you
learned probably in grade 9 to find the equation of the tangent.
#2
I would use the product rule to find the derivative
set that result equal to zero and solve for x
#3
you should get y' = (x-1)^5 (x+5)^6 (13x+23)
draw your conclusion
3) Again you need to find your derivative for your equation! Use power rule:
F'(x) = 6(x-1)^5(x+5)^7 + ((x-1)^6)(7(x+5)^6)
To get your critical values you need to set this 0 (remember the derivative is just slope so we are finding where the slope is equal to 0):
0 = 6(x-1)^5(x+5)^7 + ((x-1)^6)(7(x+5)^6)
... after a bunch of algebra and factoring... we get...
x = -23/13, x = -5, x = -1
So the answer is: D
P.S No calculator was harm in this answer (LOL)
I have a problem using the product rule
huh? just plug and chug
If u and v are functions of x, then if
y = uv
y' = u'v + uv'
you gotta problem with that, pal?
2) Again you have to use derivatives (isn't a beautiful tool!). Find the derivative of your function!
Again use power rule!
This is the derivative ==> F'(x) = 3x^2+16x+21
Now equal it to 0 to find the critical points!
0 = 3x^2+16x+21
The points are x = -7/3, x = -3
Ok now check your intervals from left of -3, in between -3 and -7/3, AND to the right of -7/3.
Now if the values decreasing in any of these intervals that means that you have maximum. If they are increasing that means that you have a minimum. Now you can just find which one is the relative maximum.
REMEMBER: Relative extrema occur where f'(x) changes sign (intervals that I mentioned).
P.S. sorry for not finishing this one... I am bit tired and I have to finish my Calculus 4 homework lol! Please guys post more integral and derivatives type of questions, they get me pumped and I love these kind of questions!!!!
Oh a problem with product rule! Tell me more I think you explain where you go wrong in there, then I will be able to help you understand it!!!
yea so whats throwing me off is how the (x+3) is squared. I don’t know if im supposed to just find the derivative without worrying about the ^2 or if i should square x+3 then find the derivative.
Umm I see...
Lets take:
f(x)=(x+2)(x+3)^2
product rule states that if your function comes in (sometimes you must a bit factoring to get there):
f(x) = u v
Then to find its derivative, you must take the derivative of the first term (leave the second term as is) and add to it (leave first term as is) the derivative of the second term as shown below
f'(x) = u' v + u v'
This is, in general from my experience, can be use in functions such as those in your problem, where it would faster to find the derivative from its original form using this rule (product rule) (REMEMBER: it only works in the form f(x) = (u)(v), so in some of my calc problem you must do some magic (**factoring and basic algebra) to get there). I just have question for you, do you want to take the derivative of this monster using factoring first:
f(x) = (3e^x + 4x^2)^2013 (e^x + lnx)^304?
It would be recommended that you do the product rule!
Please reply if you have any more questions