Find the tangent approximation for sqrt(5+x) near 3
To find the tangent approximation for sqrt(5+x) near 3, we can use the concept of linear approximation.
First, let's find the value of the function at x = 3. Evaluating sqrt(5+3):
sqrt(5+3) = sqrt(8) = 2.83 (rounded to 2 decimal places)
Next, let's find the derivative of sqrt(5+x) with respect to x.
The derivative of sqrt(5+x) can be found using the chain rule.
Let u = 5+x, then sqrt(u) = u^(1/2).
Using the chain rule, the derivative is given by:
d(sqrt(u))/du * du/dx
= (1/2)(u^(-1/2)) * 1
= 1 / (2 * sqrt(u))
Substituting back the value of u = 5 + x:
Derivative of sqrt(5+x) = 1 / (2 * sqrt(5+x))
Now, we can use the tangent approximation formula:
f(x) ≈ f(a) + f'(a)(x - a)
where f(x) is the original function, f'(a) is the derivative of the function evaluated at a, and (x - a) is the difference between x and a.
In our case, a = 3 and f(x) = sqrt(5+x).
Plugging in the values:
f(x) ≈ sqrt(8) + (1 / (2 * sqrt(8)))(x - 3)
Simplifying:
f(x) ≈ 2.83 + (1 / (2 * 2.83))(x - 3)
f(x) ≈ 2.83 + (1 / 5.66)(x - 3)
f(x) ≈ 2.83 + 0.1768(x - 3)
Therefore, the tangent approximation for sqrt(5+x) near 3 is 2.83 + 0.1768(x - 3).