How do you find the tangent equation of #f(x) = (sqrt(x) + 1)/(sqrt(x) + 2)# at a point with x=4?

Question

Isabella Ackerman · Accepted Answer

The general formula for a tangent at point #(x_o,f(x_o))# is #y-f(x_o)=f'(x_o)*(x-x_o)# Now we have that #x_o=4# ,#f(4)=3/4# and  #f'(x)=1/(2*(sqrtx+2)^2*sqrtx)# hence #f'(4)=1/64# Hence the tangent is #y-3/4=1/64*(x-3/4)# or #y=1/64*x+11/16#

Christopher Allers · Answer

undefined To find the tangent equation of f(x) = (sqrt(x) + 1)/(sqrt(x) + 2) at a point with x=4, we need to find the derivative of f(x) and evaluate it at x=4. The derivative of f(x) can be found using the quotient rule, which states that the derivative of (u/v) is (v * du/dx - u * dv/dx) / v^2.

First, let's find the derivative of the numerator, sqrt(x) + 1. The derivative of sqrt(x) is (1/2) * (x)^(-1/2), and the derivative of 1 is 0. Therefore, the derivative of the numerator is (1/2) * (x)^(-1/2).

Next, let's find the derivative of the denominator, sqrt(x) + 2. The derivative of sqrt(x) is (1/2) * (x)^(-1/2), and the derivative of 2 is 0. Therefore, the derivative of the denominator is (1/2) * (x)^(-1/2).

Now, let's apply the quotient rule. The derivative of f(x) is [(sqrt(x) + 2) * (1/2) * (x)^(-1/2) - (sqrt(x) + 1) * (1/2) * (x)^(-1/2)] / (sqrt(x) + 2)^2.

Simplifying this expression, we get [(sqrt(x) + 2) - (sqrt(x) + 1)] / 2 * (x)^(-1/2) * (sqrt(x) + 2)^2.

Further simplifying, we have [sqrt(x) + 2 - sqrt(x) - 1] / 2 * (x)^(-1/2) * (sqrt(x) + 2)^2.

This simplifies to 1 / 2 * (x)^(-1/2) * (sqrt(x) + 2)^2.

Now, let's evaluate this derivative at x=4. Plugging in x=4, we get 1 / 2 * (4)^(-1/2) * (sqrt(4) + 2)^2.

Simplifying this expression, we have 1 / 2 * (4)^(-1/2) * (2 + 2)^2.

Further simplifying, we get 1 / 2 * (4)^(-1/2) * 4^2.

This simplifies to 1 / 2 * (4)^(-1/2) * 16.

Simplifying further, we have 1 / 2 * (1/2) * 16.

This simplifies to 1 / 4 * 16.

Finally, we get 4 as the value of the derivative at x=4.

Therefore, the tangent equation of f(x) = (sqrt(x) + 1)/(sqrt(x) + 2) at the point with x=4 is y = f(4) + f'(4) * (x - 4).

Plugging in x=4 and f(4) = (sqrt(4) + 1)/(sqrt(4) + 2), we can calculate the value of f(4).

Simplifying, we get f(4) = (2 + 1)/(2 + 2).

Further simplifying, we have f(4) = 3/4.

Therefore, the tangent equation of f(x) = (sqrt(x) + 1)/(sqrt(x) + 2) at the point with x=4 is y = 3/4 + 4 * (x - 4).