equation of tangent line of the function sqrt (3x+5) where x=10
f(x) = (3x+5)^.5
f'(x) = .5 (3x+5)^-.5 * 3
= 1.5 (3x+5)^-.5
when x = 10
f(x) = sqrt(35)
f'(x) = 1.5 /sqrt(35)
y = m x + b
sqrt(35) = [1.5/sqrt(35)]10 + b
35 = 15 + b(sqrt(35))
20 /sqrt(35) = b
so
y = [1.5/sqrt(35)] x + 20 /sqrt(35)
or
sqrt(35) y = 1.5 x + 20
thanks Damon, in HS and this is kicking my butt!
You are welcome.
To find the equation of the tangent line to a function at a given point, you'll need to know the derivative of the function and the coordinates of the point.
Step 1: Find the derivative of the function:
To differentiate the function f(x) = √(3x + 5), you can use the power rule for differentiation. The power rule states that for any function of the form f(x) = x^n, the derivative is given by f'(x) = n*x^(n-1).
In this case, n=1/2, since the square root function is equivalent to raising to the power of 1/2. Hence, we have:
f'(x) = (1/2)*(3x + 5)^(-1/2)
Step 2: Substitute the value of x=10 into the derivative:
f'(10) = (1/2)*(3*10 + 5)^(-1/2) = (1/2)*(35)^(-1/2)
Step 3: Find the y-coordinate of the point on the function at x=10:
To find the y-coordinate, substitute x=10 into the original function:
f(10) = √(3*10 + 5) = √35
Now, you have the slope of the tangent line (given by f'(10)) and the point (10, √35).
Step 4: Use the point-slope form of a line to find the equation:
The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) are the coordinates of a point on the line, and m is the slope.
Plugging in the values:
y - √35 = (1/2)*(35)^(-1/2) * (x - 10)
This equation represents the tangent line to the function √(3x + 5) at the point (10, √35).