Given
[tex]$f^{\prime \prime}(x)=9 x-4$[/tex]
and [tex]$f^{\prime}(0)=2$[/tex] and [tex]$f(0)=5$[/tex].
Find [tex]$f^{\prime}(x)=[/tex] [$\square$]
and find [tex]$f(2)=[/tex] [$\square$]
Answer (2)
We determined that the first derivative of the function is f ′ ( x ) = 2 9 x 2 − 4 x + 2 and the value of the function at x = 2 is f ( 2 ) = 13 .
;
Integrate the second derivative f ′′ ( x ) = 9 x − 4 to find the first derivative: f ′ ( x ) = 2 9 x 2 − 4 x + C 1 .
Use the initial condition f ′ ( 0 ) = 2 to find C 1 = 2 , so f ′ ( x ) = 2 9 x 2 − 4 x + 2 .
Integrate the first derivative to find the function: f ( x ) = 2 3 x 3 − 2 x 2 + 2 x + C 2 .
Use the initial condition f ( 0 ) = 5 to find C 2 = 5 , so f ( x ) = 2 3 x 3 − 2 x 2 + 2 x + 5 , and evaluate f ( 2 ) = 13 .
Explanation
Problem Setup We are given the second derivative of a function, f ′′ ( x ) = 9 x − 4 , and the initial conditions f ′ ( 0 ) = 2 and f ( 0 ) = 5 . Our goal is to find the first derivative f ′ ( x ) and the value of the function at x = 2 , i.e., f ( 2 ) .
Integrating the Second Derivative First, we need to find the first derivative f ′ ( x ) by integrating the second derivative f ′′ ( x ) with respect to x :
f ′ ( x ) = ∫ f ′′ ( x ) d x = ∫ ( 9 x − 4 ) d x
Finding the First Derivative Integrating 9 x − 4 with respect to x , we get: f ′ ( x ) = 2 9 x 2 − 4 x + C 1
Applying the Initial Condition for the First Derivative Now, we use the initial condition f ′ ( 0 ) = 2 to find the constant of integration C 1 :
2 = 2 9 ( 0 ) 2 − 4 ( 0 ) + C 1 C 1 = 2 So, the first derivative is: f ′ ( x ) = 2 9 x 2 − 4 x + 2
Integrating the First Derivative Next, we find the function f ( x ) by integrating the first derivative f ′ ( x ) with respect to x :
f ( x ) = ∫ f ′ ( x ) d x = ∫ ( 2 9 x 2 − 4 x + 2 ) d x
Finding the Function Integrating 2 9 x 2 − 4 x + 2 with respect to x , we get: f ( x ) = 2 9 ⋅ 3 x 3 − 4 ⋅ 2 x 2 + 2 x + C 2 f ( x ) = 2 3 x 3 − 2 x 2 + 2 x + C 2
Applying the Initial Condition for the Function Now, we use the initial condition f ( 0 ) = 5 to find the constant of integration C 2 :
5 = 2 3 ( 0 ) 3 − 2 ( 0 ) 2 + 2 ( 0 ) + C 2 C 2 = 5 So, the function is: f ( x ) = 2 3 x 3 − 2 x 2 + 2 x + 5
Evaluating f(2) Finally, we evaluate f ( 2 ) :
f ( 2 ) = 2 3 ( 2 ) 3 − 2 ( 2 ) 2 + 2 ( 2 ) + 5 f ( 2 ) = 2 3 ( 8 ) − 2 ( 4 ) + 4 + 5 f ( 2 ) = 12 − 8 + 4 + 5 f ( 2 ) = 13
Final Answer Therefore, we have found that f ′ ( x ) = 2 9 x 2 − 4 x + 2 and f ( 2 ) = 13 .
Examples
In physics, if f ′′ ( x ) represents the acceleration of an object, then f ′ ( x ) is its velocity, and f ( x ) is its position. Given the acceleration function and initial velocity and position, we can determine the object's velocity at any time x and its position at any time x . This is a fundamental concept in classical mechanics.
