Find the equation of the tangent line to the graph of f(x) = 6x ln(x + 4) at x = 2.
dy/dx = 6x d/dx [ln(x+4)]+ 6 ln(x+4)
= 6 x /(x+4) + 6 ln(x+4)
so at x = 2
dy/dx = slope = m = 12/ln(6) + 6 ln (6)
but ln (6) = 1.79
so use that to find the slope m
then we have
y = m x + b and we know m
find y at x = 2
y= 12 (1.79) = 21.5
so use that
21.5 = 2 m + b
we know m, solve for b
the end
Thank you!
You are welcome.
To find the equation of the tangent line to the graph of f(x), we need the slope of the tangent line and a point on the line.
1. To find the slope of the tangent line, we can use the derivative of the function f(x). Take the derivative of f(x) with respect to x.
f'(x) = (d/dx)(6x ln(x + 4))
To differentiate this function, we can use the product rule:
f'(x) = 6(ln(x + 4) + x/(x + 4))
2. Now we have the expression for the slope of the tangent line at any x-value. To find the slope at x = 2, substitute x = 2 into f'(x):
slope = f'(2) = 6(ln(2 + 4) + 2/(2 + 4))
3. Simplify the expression to find the precise value of the slope.
slope = 6(ln(6) + 2/6) = 6(ln(6) + 1/3)
4. Next, we need a point on the tangent line. We can use the point (2, f(2)) since x = 2 is the point where we want to find the tangent line.
f(2) = 6(2 ln(2 + 4)) = 6(2 ln(6))
Now we have a point on the tangent line: (2, 12 ln(6))
5. Finally, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point on the line.
Using the point (2, 12 ln(6)) and the slope 6(ln(6) + 1/3), the equation of the tangent line is:
y - (12 ln(6)) = 6(ln(6) + 1/3)(x - 2)
Simplifying this equation will give you the final equation of the tangent line.