Post

Created by @nathanedwards
 at October 31st 2023, 7:28:39 pm.

AP Calculus AB Exam Question:

Let f(x)=xf(x) = \sqrt{x}, where xx is a positive number.

(a) Find the linear approximation of f(x)f(x) near x=4x = 4.

(b) Use the linear approximation from part (a) to estimate 4.2\sqrt{4.2}.

Answer:

(a) To find the linear approximation of f(x)f(x) near x=4x = 4, we will use the concept of tangent lines and the differential of f(x)f(x).

Recall that the equation of the tangent line to a curve at a given point (a,f(a))(a, f(a)) is given by:

yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a)

Taking the derivative of f(x)f(x), we have:

f(x)=12xf'(x) = \frac{1}{2\sqrt{x}}

Now, let's find the equation of the tangent line to f(x)f(x) at x=4x = 4. In this case, a=4a = 4 and f(a)=f(4)=4=2f(a) = f(4) = \sqrt{4} = 2. Plugging these values into the equation of the tangent line, we have:

y2=124(x4)y - 2 = \frac{1}{2\sqrt{4}}(x - 4)

Simplifying further, we get:

y=2+14(x4)y = 2 + \frac{1}{4}(x - 4)

Therefore, the linear approximation of f(x)f(x) near x=4x = 4 is L(x)=2+14(x4)L(x) = 2 + \frac{1}{4}(x - 4).

(b) To estimate 4.2\sqrt{4.2} using the linear approximation above, we need to evaluate L(4.2)L(4.2).

Plugging x=4.2x = 4.2 into the linear approximation equation, we have:

L(4.2)=2+14(4.24)L(4.2) = 2 + \frac{1}{4}(4.2 - 4)

Simplifying, we get:

L(4.2)=2+14(0.2)=2+14(0.2)=2+0.05=2.05L(4.2) = 2 + \frac{1}{4}(0.2) = 2 + \frac{1}{4}(0.2) = 2 + 0.05 = 2.05

Therefore, 4.2\sqrt{4.2} is approximately equal to 2.05 using the linear approximation.