$f(x)=3 x^2+5 x+6$
(i) $f^{\prime}(0)$
(ii) $f^{\prime}(-2)$
(iii) $f^{\prime \prime}(1)$
Answer (2)
The first derivative of the function at x = 0 is 5, at x = − 2 is -7, and the second derivative at x = 1 is 6. The calculations involve differentiation using the power rule. Overall, these derivatives help us understand the function's behavior at the specified points.
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Find the first derivative: f ′ ( x ) = 6 x + 5 .
Find the second derivative: f ′′ ( x ) = 6 .
Evaluate f ′ ( 0 ) : f ′ ( 0 ) = 5 .
Evaluate f ′ ( − 2 ) : f ′ ( − 2 ) = − 7 .
Evaluate f ′′ ( 1 ) : f ′′ ( 1 ) = 6 .
f ′ ( 0 ) = 5 , f ′ ( − 2 ) = − 7 , f ′′ ( 1 ) = 6
Explanation
Problem Analysis We are given the function f ( x ) = 3 x 2 + 5 x + 6 and we need to find f ′ ( 0 ) , f ′ ( − 2 ) , and f ′′ ( 1 ) . This involves finding the first and second derivatives of the function and then evaluating them at the specified points.
Finding the First Derivative First, let's find the first derivative, f ′ ( x ) . Using the power rule, we have:
f ( x ) = 3 x 2 + 5 x + 6
f ′ ( x ) = 3 ( 2 x ) + 5 ( 1 ) + 0
f ′ ( x ) = 6 x + 5
Finding the Second Derivative Now, let's find the second derivative, f ′′ ( x ) . Taking the derivative of f ′ ( x ) , we get:
f ′ ( x ) = 6 x + 5
f ′′ ( x ) = 6
Evaluating the Derivatives Now we can evaluate f ′ ( 0 ) , f ′ ( − 2 ) , and f ′′ ( 1 ) .
f ′ ( 0 ) = 6 ( 0 ) + 5 = 5
f ′ ( − 2 ) = 6 ( − 2 ) + 5 = − 12 + 5 = − 7
f ′′ ( 1 ) = 6
Examples
Understanding derivatives is crucial in many real-world applications. For example, in physics, if f ( x ) represents the position of an object at time x , then f ′ ( x ) represents the velocity of the object, and f ′′ ( x ) represents the acceleration. Evaluating these at specific times gives us the instantaneous velocity and acceleration at those moments. This is also used in economics to analyze marginal cost and revenue, helping businesses make informed decisions.
