Consider the following function.

[tex]f(x)=\frac{21}{x^2+3}[/tex]

Find the first and second derivatives.

[tex]\begin{array}{l}
f^{\prime}(x)=\square \\
f^{\prime}(x)=\square
\end{array}[/tex]

Find any values of [tex]c[/tex] such that [tex]f^{\prime \prime}(c)=0[/tex]. (Enter your answer as a comma-separated list. If any answer does not exist, enter DNE.)

[tex]c=\square[/tex]

Determine the open intervals on which the graph of the function is concave upward or concave downward. (Enter your answers using interval notation. If an answer does not exist, enter DNE.)

concave upward [tex]\square[/tex]
concave downward [tex]\square[/tex]

Find the first derivative: f ′ ( x ) = ( x 2 + 3 ) 2 − 42 x ​ .
Find the second derivative: f ′′ ( x ) = ( x 2 + 3 ) 3 126 ( x 2 − 1 ) ​ .
Solve f ′′ ( c ) = 0 to find c = − 1 , 1 .
Determine concavity: Concave upward on ( − ∞ , − 1 ) and ( 1 , ∞ ) , concave downward on ( − 1 , 1 ) .
f ′ ( x ) = ( x 2 + 3 ) 2 − 42 x ​ , f ′′ ( x ) = ( x 2 + 3 ) 3 126 ( x 2 − 1 ) ​ , c = − 1 , 1 , concave upward: ( − ∞ , − 1 ) ∪ ( 1 , ∞ ) , concave downward: ( − 1 , 1 ) ​

Explanation

Problem Analysis We are given the function f ( x ) = x 2 + 3 21 ​ . Our goal is to find the first and second derivatives, determine the values of c for which f ′′ ( c ) = 0 , and find the intervals where the graph of f ( x ) is concave upward or concave downward.

Finding the First Derivative First, we find the first derivative f ′ ( x ) . Using the quotient rule, we have: f ′ ( x ) = ( x 2 + 3 ) 2 ( 0 ) ( x 2 + 3 ) − ( 21 ) ( 2 x ) ​ = ( x 2 + 3 ) 2 − 42 x ​ So, f ′ ( x ) = ( x 2 + 3 ) 2 − 42 x ​ .

Finding the Second Derivative Next, we find the second derivative f ′′ ( x ) . Using the quotient rule again on f ′ ( x ) , we get: f ′′ ( x ) = ( x 2 + 3 ) 4 ( − 42 ) (( x 2 + 3 ) 2 ) − ( − 42 x ) ( 2 ( x 2 + 3 ) ( 2 x )) ​ f ′′ ( x ) = ( x 2 + 3 ) 4 − 42 ( x 2 + 3 ) 2 + 168 x 2 ( x 2 + 3 ) ​ f ′′ ( x ) = ( x 2 + 3 ) 3 − 42 ( x 2 + 3 ) + 168 x 2 ​ f ′′ ( x ) = ( x 2 + 3 ) 3 − 42 x 2 − 126 + 168 x 2 ​ f ′′ ( x ) = ( x 2 + 3 ) 3 126 x 2 − 126 ​ = ( x 2 + 3 ) 3 126 ( x 2 − 1 ) ​ So, f ′′ ( x ) = ( x 2 + 3 ) 3 126 ( x 2 − 1 ) ​ .

Finding Inflection Points Now, we find the values of c such that f ′′ ( c ) = 0 . We set f ′′ ( c ) = 0 :
( c 2 + 3 ) 3 126 ( c 2 − 1 ) ​ = 0 This is true when c 2 − 1 = 0 , which means c 2 = 1 , so c = ± 1 .

Determining Concavity Intervals Next, we determine the intervals where the graph of f ( x ) is concave upward or concave downward. We analyze the sign of f ′′ ( x ) :



If x < − 1 , then 1"> x 2 > 1 , so 0"> x 2 − 1 > 0 , and 0"> f ′′ ( x ) > 0 . Thus, f ( x ) is concave upward on ( − ∞ , − 1 ) .
If − 1 < x < 1 , then x 2 < 1 , so x 2 − 1 < 0 , and f ′′ ( x ) < 0 . Thus, f ( x ) is concave downward on ( − 1 , 1 ) .
If 1"> x > 1 , then 1"> x 2 > 1 , so 0"> x 2 − 1 > 0 , and 0"> f ′′ ( x ) > 0 . Thus, f ( x ) is concave upward on ( 1 , ∞ ) .


Final Answer Therefore, the function is concave upward on the intervals ( − ∞ , − 1 ) and ( 1 , ∞ ) , and concave downward on the interval ( − 1 , 1 ) . The values of c such that f ′′ ( c ) = 0 are c = − 1 , 1 .

Examples
Understanding concavity is crucial in various fields. For instance, in economics, it helps analyze the rate of change of cost functions. If a cost function is concave downward, it indicates decreasing marginal costs, which can inform business decisions about production levels and pricing strategies. Similarly, in physics, concavity helps analyze potential energy curves to determine stability of systems.

Answered by GinnyAnswer | 2025-07-05

The first derivative of the function is f ′ ( x ) = ( x 2 + 3 ) 2 − 42 x ​ and the second derivative is f ′′ ( x ) = ( x 2 + 3 ) 3 126 ( x 2 − 1 ) ​ . Values of c such that f ′′ ( c ) = 0 are − 1 , 1 , with concave upward intervals at ( − ∞ , − 1 ) ∪ ( 1 , ∞ ) and concave downward at ( − 1 , 1 ) .
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Answered by Anonymous | 2025-07-11