Consider the following function:

[tex]f(x)=x^2-3 x+9[/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 ) = 2 x − 3 .
Find the second derivative: f ′′ ( x ) = 2 .
Determine values of c where f ′′ ( c ) = 0 : Since f ′′ ( x ) = 2 is a constant, there are no values of c such that f ′′ ( c ) = 0 , so c = D NE .
Determine concavity: The function is concave upward on ( − ∞ , ∞ ) and never concave downward, so concave downward is D NE .

f ′ ( x ) = 2 x − 3 , f ′′ ( x ) = 2 , c = D NE , concave upward ( − ∞ , ∞ ) , concave downward D NE ​
Explanation

Problem Analysis We are given the function f ( x ) = x 2 − 3 x + 9 . Our goal is to find the first derivative f ′ ( x ) , the second derivative f ′′ ( x ) , the value(s) of c such that f ′′ ( c ) = 0 , and the intervals where the function is concave upward and concave downward.

Finding the First Derivative First, we find the first derivative of f ( x ) using the power rule. The power rule states that if f ( x ) = x n , then f ′ ( x ) = n x n − 1 . Applying this rule, we get:


f ′ ( x ) = d x d ​ ( x 2 − 3 x + 9 ) = 2 x − 3

Finding the Second Derivative Next, we find the second derivative by differentiating the first derivative:

f ′′ ( x ) = d x d ​ ( 2 x − 3 ) = 2

Finding Values Where Second Derivative is Zero Now, we need to find the value(s) of c such that f ′′ ( c ) = 0 . Since f ′′ ( x ) = 2 for all x , there is no value of c for which f ′′ ( c ) = 0 . Therefore, the answer is DNE.

Determining Concave Upward Intervals To determine the intervals where the function is concave upward, we need to find where 0"> f ′′ ( x ) > 0 . Since f ′′ ( x ) = 2 , which is always greater than 0, the function is concave upward for all x . Therefore, the interval where the function is concave upward is ( − ∞ , ∞ ) .

Determining Concave Downward Intervals To determine the intervals where the function is concave downward, we need to find where f ′′ ( x ) < 0 . Since f ′′ ( x ) = 2 , which is never less than 0, the function is never concave downward. Therefore, the interval where the function is concave downward is DNE.

Final Answer In summary:


f ′ ( x ) = 2 x − 3 f ′′ ( x ) = 2 c = D NE Concave upward: ( − ∞ , ∞ ) Concave downward: DNE
Examples
Understanding concavity is crucial in various fields. For example, in economics, it helps analyze cost functions. If a cost function is concave upward, it indicates increasing marginal costs, meaning each additional unit costs more to produce than the previous one. Conversely, concave downward indicates decreasing marginal costs, which can inform decisions about scaling production.

Answered by GinnyAnswer | 2025-07-05

The first derivative of the function is f ′ ( x ) = 2 x − 3 and the second derivative is f ′′ ( x ) = 2 . There are no values of c such that f ′′ ( c ) = 0 , so c = D NE . The function is concave upward on ( − ∞ , ∞ ) and never concave downward (DNE).
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Answered by Anonymous | 2025-07-06