Using the definition of the derivative, find $f^{\prime}(x)$. Then find $f^{\prime}(1), f^{\prime}(2)$, and $f^{\prime}(3)$ when the derivative exists.
$f(x)=-x^2+2 x-2$
$f^{\prime}(x)=$ $\square$
(Type an expression using $x$ as the variable.)
Select the correct answer below and, if necessary, fill in the answer box to complete your choice.
A. $f(1)=$ $\square$
(Type an integer or a simplified fraction.)
B. The derivative does not exist.
Select the correct answer below and, if necessary, fill in the answer box to complete your choice.
A. $f^{\prime}(2)=$ $\square$
(Type an integer or a simplified fraction.)
B. The derivative does not exist.
Select the correct answer below and, if necessary, fill in the answer box to complete your choice.
A. $f^{\prime}(3)=$ $\square$
(Type an integer or a simplified fraction.)
B. The derivative does not exist.
Answer (1)
Find f ( x + h ) : Substitute x + h into f ( x ) to get f ( x + h ) = − x 2 − 2 x h − h 2 + 2 x + 2 h − 2 .
Calculate h f ( x + h ) − f ( x ) : Simplify the expression to get h f ( x + h ) − f ( x ) = − 2 x − h + 2 .
Find the limit as h → 0 : Take the limit of the simplified expression as h approaches 0, resulting in f ′ ( x ) = − 2 x + 2 .
Evaluate f ′ ( 1 ) , f ′ ( 2 ) , and f ′ ( 3 ) : Substitute x = 1 , 2 , 3 into f ′ ( x ) to get f ′ ( 1 ) = 0 , f ′ ( 2 ) = − 2 , and f ′ ( 3 ) = − 4 .
f ′ ( x ) = − 2 x + 2 , f ′ ( 1 ) = 0 , f ′ ( 2 ) = − 2 , f ′ ( 3 ) = − 4
Explanation
Problem Analysis We are given the function f ( x ) = − x 2 + 2 x − 2 . Our goal is to find the derivative f ′ ( x ) using the definition of the derivative, and then evaluate the derivative at x = 1 , 2 , 3 .
Definition of Derivative The definition of the derivative is given by: f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) We will use this definition to find the derivative of the given function.
Finding f(x+h) First, we need to find f ( x + h ) . We substitute x + h into the function f ( x ) :
f ( x + h ) = − ( x + h ) 2 + 2 ( x + h ) − 2 Expanding this, we get: f ( x + h ) = − ( x 2 + 2 x h + h 2 ) + 2 x + 2 h − 2 f ( x + h ) = − x 2 − 2 x h − h 2 + 2 x + 2 h − 2
Finding f(x+h) - f(x) Now, we find f ( x + h ) − f ( x ) :
f ( x + h ) − f ( x ) = ( − x 2 − 2 x h − h 2 + 2 x + 2 h − 2 ) − ( − x 2 + 2 x − 2 ) f ( x + h ) − f ( x ) = − x 2 − 2 x h − h 2 + 2 x + 2 h − 2 + x 2 − 2 x + 2 f ( x + h ) − f ( x ) = − 2 x h − h 2 + 2 h
Dividing by h Next, we divide by h :
h f ( x + h ) − f ( x ) = h − 2 x h − h 2 + 2 h h f ( x + h ) − f ( x ) = − 2 x − h + 2
Finding the Limit Now, we take the limit as h approaches 0: f ′ ( x ) = h → 0 lim ( − 2 x − h + 2 ) f ′ ( x ) = − 2 x + 2
Evaluating the Derivative Now we evaluate f ′ ( 1 ) , f ′ ( 2 ) , and f ′ ( 3 ) :
f ′ ( 1 ) = − 2 ( 1 ) + 2 = 0 f ′ ( 2 ) = − 2 ( 2 ) + 2 = − 4 + 2 = − 2 f ′ ( 3 ) = − 2 ( 3 ) + 2 = − 6 + 2 = − 4
Final Answer Therefore, f ′ ( x ) = − 2 x + 2 , f ′ ( 1 ) = 0 , f ′ ( 2 ) = − 2 , and f ′ ( 3 ) = − 4 .
Examples
Understanding derivatives is crucial in many real-world applications. For instance, in physics, if f ( x ) represents the position of an object at time x , then f ′ ( x ) gives the velocity of the object at time x . Similarly, in economics, if f ( x ) represents the cost of producing x items, then f ′ ( x ) gives the marginal cost, which is the approximate cost of producing one more item. This concept helps businesses make informed decisions about production levels and pricing strategies.
