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In Mathematics / College | 2025-07-04

$f(x)=3 x^2+5 x+6$

(i) $f^{\prime}(0)$ (ii) $f^{\prime \prime}(1)$

Asked by successgodday14

Answer (2)

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 .
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Answered by Anonymous | 2025-07-04

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.

Answered by GinnyAnswer | 2025-07-04