Question

question 3
let ( f(x) = sqrt{x} ). it can be seen using the definition of the derivative that ( f(25) = \frac{1}{10} ). use the equation of the line tangent to the graph of ( f(x) ) at ( x = 25 ) to approximate ( sqrt{25.2} ).
( \bigcirc 5 )
( \bigcirc 5.02 )
( \bigcirc 0 )
( \bigcirc 5.01996 )
( \bigcirc 5.01 )

Explanation:

Step1: Recall the linear approximation formula

The linear approximation of a function \( f(x) \) at \( x = a \) is given by \( L(x)=f(a)+f^{\prime}(a)(x - a) \). Here, \( f(x)=\sqrt{x} \), \( a = 25 \), and we want to approximate \( f(25.2) \), so \( x=25.2 \).

Step2: Calculate \( f(a) \)

First, find \( f(25) \). Since \( f(x)=\sqrt{x} \), then \( f(25)=\sqrt{25}=5 \).

Step3: Identify \( f^{\prime}(a) \)

We are given that \( f^{\prime}(25)=\frac{1}{10} \).

Step4: Apply the linear approximation formula

Substitute \( a = 25 \), \( f(a)=5 \), \( f^{\prime}(a)=\frac{1}{10} \), and \( x = 25.2 \) into the linear approximation formula \( L(x)=f(a)+f^{\prime}(a)(x - a) \).
So, \( L(25.2)=f(25)+f^{\prime}(25)(25.2 - 25) \)
\(=5+\frac{1}{10}(0.2) \)
\(=5 + 0.02 \)
\(=5.02 \)

Answer:

5.02