The first derivative of the function f ( x ) = 3 x 2 + 5 x + 6 evaluated at x = 0 is 5 , and the second derivative evaluated at x = 1 is 6 . Therefore, f ′ ( 0 ) = 5 and f ′′ ( 1 ) = 6 .
;
Find the first derivative f ′ ( x ) of the function f ( x ) = 3 x 2 + 5 x + 6 using the power rule: f ′ ( x ) = 6 x + 5 .
Evaluate the first derivative at x = 0 : f ′ ( 0 ) = 6 ( 0 ) + 5 = 5 .
Find the second derivative f ′′ ( x ) by differentiating f ′ ( x ) : f ′′ ( x ) = 6 .
Evaluate the second derivative at x = 1 : f ′′ ( 1 ) = 6 . Thus, the final answers are 5 and 6 .
Explanation
Problem Analysis We are given the function f ( x ) = 3 x 2 + 5 x + 6 and asked to find the values of its first derivative at x = 0 , denoted as f ′ ( 0 ) , and its second derivative at x = 1 , denoted as f ′′ ( 1 ) .
Finding the First Derivative First, we need to find the first derivative of f ( x ) with respect to x . Using the power rule, which states that d x d ( x n ) = n x n − 1 , we have:
f ′ ( x ) = d x d ( 3 x 2 + 5 x + 6 ) = 3 ( 2 x ) + 5 ( 1 ) + 0 = 6 x + 5
Evaluating the First Derivative at x=0 Next, we evaluate f ′ ( x ) at x = 0 :
f ′ ( 0 ) = 6 ( 0 ) + 5 = 0 + 5 = 5
Finding the Second Derivative Now, we find the second derivative of f ( x ) with respect to x . This is the derivative of f ′ ( x ) :
f ′′ ( x ) = d x d ( 6 x + 5 ) = 6 ( 1 ) + 0 = 6
Evaluating the Second Derivative at x=1 Finally, we evaluate f ′′ ( x ) at x = 1 :
f ′′ ( 1 ) = 6
Since the second derivative is a constant, its value is the same for all x .
Final Answer Therefore, f ′ ( 0 ) = 5 and f ′′ ( 1 ) = 6 .
Examples
Understanding derivatives is crucial in physics, especially when analyzing motion. For example, if f ( x ) represents the position of an object at time x , then f ′ ( x ) gives the object's velocity, and f ′′ ( x ) gives its acceleration. Evaluating these derivatives at specific times helps us understand the object's motion at those instants. This is fundamental in fields like mechanics and aerospace engineering.