Asked by Morris
Hellp I have a calculus test on Monday
2. Calculate the slope of the tangent to the given function at the given point or value of x
a. f(x)=3/x+1,P(2,1)
b.h(x)=2/squareroot x + 5, P(4,2/3)
****Full solutions to please because I do not know what I'm doing and so confused with calculus
2. Calculate the slope of the tangent to the given function at the given point or value of x
a. f(x)=3/x+1,P(2,1)
b.h(x)=2/squareroot x + 5, P(4,2/3)
****Full solutions to please because I do not know what I'm doing and so confused with calculus
Answers
Answered by
Reiny
2a)
I will assume you mean f(x) = 3/(x+1)
I would write it f(x) = 3(x+1)^-1
then f ' (x) = -3(x+1)^-2
= -3/(x+1)^2
at P(2,1), slope = -3/(3)^2 = -3/9 = -1/3
h(x) = 2/√(x+5)
= 2(x+5)^(-1/2)
h ' (x) = -1(x+5)^(-3/2)
= -1/( √(x+5) )^3
you finish it
I will assume you mean f(x) = 3/(x+1)
I would write it f(x) = 3(x+1)^-1
then f ' (x) = -3(x+1)^-2
= -3/(x+1)^2
at P(2,1), slope = -3/(3)^2 = -3/9 = -1/3
h(x) = 2/√(x+5)
= 2(x+5)^(-1/2)
h ' (x) = -1(x+5)^(-3/2)
= -1/( √(x+5) )^3
you finish it
Answered by
Morris
how you get f ' (x) = -3(x+1)^-2 what numbers do i multiply with
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.