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
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
how you get f ' (x) = -3(x+1)^-2 what numbers do i multiply with
No problem! I'm here to help you understand calculus better and solve these problems.
To find the slope of the tangent to a function at a given point, you need to first take the derivative of the function, and then substitute the x-coordinate of the given point into the derivative to find the slope.
Let's start with problem (a):
a. f(x) = 3/(x + 1), P(2, 1)
Step 1: Finding the derivative of f(x)
To find the derivative of the function f(x), you need to use the power rule or the quotient rule. In this case, you can use the quotient rule since you have a fraction.
The quotient rule states:
If f(x) = u(x)/v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x))/(v(x))^2
Using the quotient rule, we can find the derivative of f(x):
f'(x) = [(3)'(x + 1) - 3(x + 1)']/(x + 1)^2
f'(x) = [0 - 3]/(x + 1)^2
f'(x) = -3/(x + 1)^2
Step 2: Evaluate f'(x) at x = 2
To find the slope of the tangent at the point P(2, 1), substitute the x-coordinate (2) into f'(x):
f'(2) = -3/(2 + 1)^2
f'(2) = -3/3^2
f'(2) = -3/9
f'(2) = -1/3
Therefore, the slope of the tangent to the function f(x) = 3/(x + 1) at the point P(2, 1) is -1/3.
Now, let's move on to problem (b):
b. h(x) = 2/√x + 5, P(4, 2/3)
Step 1: Finding the derivative of h(x)
Again, we need to find the derivative of the function h(x) using the quotient rule.
h'(x) = [(2)'(√x + 5) - 2(√x + 5)']/(√x + 5)^2
h'(x) = [0 - 2(1/2√x)]/(√x + 5)^2
h'(x) = -1/(√x + 5)^2√x
Step 2: Evaluate h'(x) at x = 4
Substitute the x-coordinate (4) into h'(x):
h'(4) = -1/(√4 + 5)^2√4
h'(4) = -1/(2 + 5)^2·2
h'(4) = -1/(7)^2·2
h'(4) = -1/(49)·2
h'(4) = -2/49
Therefore, the slope of the tangent to the function h(x) = 2/√x + 5 at the point P(4, 2/3) is -2/49.
I hope this helps you understand how to solve these types of problems in calculus. Good luck with your test! Let me know if you have any more questions.