If [tex]\frac{dy}{dx} \big|_{x=2} = 2[/tex], then [tex]f'(2) = ?[/tex]
A. [tex]\lim_{h \to 0} \frac{f(h) - f(2)}{h}[/tex]
B. [tex]\lim_{h \to 0} \frac{f(2) - f(h)}{h}[/tex]
C. [tex]\lim_{h \to 0} \frac{f(2-h) - f(2)}{h}[/tex]
D. [tex]\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}[/tex]
Answer (1)
To determine which option correctly represents f ′ ( 2 ) , we need to understand the concept of a derivative. The derivative of a function f at a specific point x = a is defined by the limit:
f ′ ( a ) = h → 0 lim h f ( a + h ) − f ( a )
This limit represents the slope of the tangent line to the function f at the point x = a . It measures how the function f changes as x changes by a small amount h .
Given that d x d y x = 2 = 2 , this means that the derivative of the function (let's assume y = f ( x ) ) at x = 2 is 2, or in other words, f ′ ( 2 ) = 2 .
Now, let's examine the options provided to determine which one corresponds to this definition of the derivative:
A. lim h → 0 h f ( h ) − f ( 2 ) - This is \textbf{not} correct because it represents the slope considering x = 0 , not x = 2 .
B. lim h → 0 h f ( 2 ) − f ( h ) - This is \textbf{not} correct because it is essentially the negative of the correct derivative definition; it compares changes from h to 2 backwards.
C. lim h → 0 h f ( 2 − h ) − f ( 2 ) - This also represents a backward difference quotient and is \textbf{not} the conventional definition for a derivative at x = 2 .
D. lim h → 0 h f ( 2 + h ) − f ( 2 ) - This correctly represents the limit definition of the derivative as it calculates the change at x = 2 by considering x = 2 + h . Therefore, this is the \textbf{correct} option matching the definition of f ′ ( 2 ) .
Therefore, the correct choice is Option D .
