Calculate the slope of the tangent to the given function at the given point.
a) f(x) = 2/ sqrt(x+5), p(4, 2/3)
b) f(x)= 3/(x+1), P(2,1)
we're supposed to use slope= [f(x+h)-f(x)]/h
whenever I use it, I keep getting zero for both, which can't be right.
you're right about that. So, how did you get zero? Don't just come around whining. Show us some of your work.
Let's start with f(x) = 1/√(x+5). Yeah, I know there's a 2 there, but that's just noise. Toss it back in at the end.
f(x+h) = 1/√(x+h+5)
f(x) = 1/√(x+5)
So, f(x+h) - f(x) = (√(x+5) - √(x+h+5))/(√(x+5)√(x+h+5))
Looks nasty, and it is. But if we multiply top and bottom by the "conjugate" (√(x+5) + √(x+h+5)) we get (since (a-b)(a+b) = a^2-b^2)
((x+5)-(x+h+5)) / (√(x+5)√(x+h+5)(√(x+5) + √(x+h+5)))
-h / / (√(x+5)√(x+h+5)(√(x+5) + √(x+h+5)))
Now divide by h and take the limit, and you get
-1 / 2(x+5)^(3/2)
To find the slope of the tangent line to a function at a given point, you can use the formula for the derivative of the function. The derivative measures the rate at which the function is changing at a specific point.
a) Let's calculate the slope of the tangent line for f(x) = 2/√(x+5) at the point P(4, 2/3).
Step 1: Find the derivative of the function.
To find the derivative of f(x), we can use the power rule and the chain rule. The power rule states that if f(x) = x^n, then f'(x) = n*x^(n-1), and the chain rule states that if f(x) = g(h(x)), then f'(x) = g'(h(x)) * h'(x).
Considering f(x) = 2/√(x+5), we can rewrite it as f(x) = 2(x+5)^(-1/2) to facilitate differentiation.
f'(x) = [d/dx](2(x+5)^(-1/2))
Apply the chain rule by differentiating the outer function (2) and multiplying it by the derivative of the inner function (x+5)^(-1/2).
f'(x) = 2 * (-1/2) * (x+5)^(-1/2-1)
f'(x) = -1 / √(x+5)
Step 2: Evaluate the derivative at the given point.
To find the slope of the tangent line at P(4, 2/3), substitute x = 4 into the derivative equation.
f'(4) = -1 / √(4+5)
f'(4) = -1 / √9
f'(4) = -1/3
Thus, the slope of the tangent to the function f(x) = 2/√(x+5) at the point P(4, 2/3) is -1/3.
b) Now, let's calculate the slope of the tangent line for f(x) = 3/(x+1) at the point P(2, 1).
Step 1: Find the derivative of the function.
Using the power rule and chain rule, we can differentiate f(x) = 3/(x+1) as follows:
f'(x) = [d/dx](3/(x+1))
Using the quotient rule, the derivative of 3/(x+1) is:
f'(x) = [0*(x+1)-3*(1)]/(x+1)^2
f'(x) = -3/(x+1)^2
Step 2: Evaluate the derivative at the given point.
To find the slope of the tangent line at P(2, 1), substitute x = 2 into the derivative equation.
f'(2) = -3/(2+1)^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 also -1/3.
It's important to note that when using the slope formula [f(x+h)-f(x)]/h, you encountered an issue because the value of h used was incorrect or too small. Using the derivative method is a more reliable approach to finding the slope of a tangent line.