Consider the following function:
[tex]y=9 x-5 \tan (x), \quad\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex]
Find the first and second derivatives.
[tex]\begin{array}{l}
y^{\prime}(x)=\square \\
y^{\prime}(x)=\square
\end{array}[/tex]
Find any values of [tex]c[/tex] such that [tex]y^{\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]
Answer (2)
Find the first derivative: y ′ ( x ) = 9 − 5 sec 2 ( x ) .
Find the second derivative: y ′′ ( x ) = − 10 sec 2 ( x ) tan ( x ) .
Solve y ′ ( c ) = 0 to find critical points: c = ± arccos ( 3 5 ) .
Determine concavity: Concave up on ( − 2 π , 0 ) and concave down on ( 0 , 2 π ) .
y ′ ( x ) = 9 − 5 sec 2 ( x ) , y ′′ ( x ) = − 10 sec 2 ( x ) tan ( x ) , c = arccos ( 3 5 ) , c = − arccos ( 3 5 ) , concave up: ( − 2 π , 0 ) , concave down: ( 0 , 2 π )
Explanation
Problem Analysis We are given the function y = 9 x − 5 tan ( x ) and asked to find its first and second derivatives, the values of c where y ′ ( c ) = 0 , and the intervals of concavity. Let's break this down step by step.
Finding the First Derivative First, we need to find the first derivative, y ′ ( x ) . Recall that the derivative of x is 1 and the derivative of tan ( x ) is sec 2 ( x ) . Therefore, using the power rule and the constant multiple rule, we have:
y ′ ( x ) = 9 − 5 sec 2 ( x )
Finding the Second Derivative Next, we find the second derivative, y ′′ ( x ) , by differentiating y ′ ( x ) . The derivative of a constant is 0. The derivative of sec 2 ( x ) is 2 sec ( x ) ⋅ sec ( x ) tan ( x ) = 2 sec 2 ( x ) tan ( x ) . Thus,
y ′′ ( x ) = − 10 sec 2 ( x ) tan ( x )
Finding Critical Points Now, we need to find the values of c such that y ′ ( c ) = 0 . This means solving the equation 9 − 5 sec 2 ( c ) = 0 for c . This is equivalent to sec 2 ( c ) = 5 9 , so cos 2 ( c ) = 9 5 , and cos ( c ) = ± 3 5 . Since c is in the interval ( − 2 π , 2 π ) , we can find the values of c using the inverse cosine function:
c = arccos ( 3 5 ) ≈ 0.7297 c = − arccos ( 3 5 ) ≈ − 0.7297
Determining Concavity Intervals Finally, we determine the intervals where the function is concave upward ( 0"> y ′′ ( x ) > 0 ) and concave downward ( y ′′ ( x ) < 0 ). Since sec 2 ( x ) is always positive, the sign of y ′′ ( x ) depends on the sign of tan ( x ) .
If 0"> tan ( x ) > 0 , then y ′′ ( x ) < 0 , so the function is concave downward. This occurs when 0 < x < 2 π .
If tan ( x ) < 0 , then 0"> y ′′ ( x ) > 0 , so the function is concave upward. This occurs when − 2 π < x < 0 .
Therefore, the function is concave upward on the interval ( − 2 π , 0 ) and concave downward on the interval ( 0 , 2 π ) .
Final Answer In summary:
First derivative: y ′ ( x ) = 9 − 5 sec 2 ( x ) Second derivative: y ′′ ( x ) = − 10 sec 2 ( x ) tan ( x ) Values of c such that y ′ ( c ) = 0 : c = arccos ( 3 5 ) , c = − arccos ( 3 5 ) Concave upward: ( − 2 π , 0 ) Concave downward: ( 0 , 2 π )
Examples
Understanding concavity is crucial in many real-world applications. For example, engineers designing bridges need to consider the concavity of the bridge's structure to ensure stability and minimize stress. Similarly, in economics, understanding the concavity of cost functions helps businesses make optimal decisions about production levels. In physics, the trajectory of a projectile can be analyzed using concavity to determine its maximum height and range.
The first derivative of the function is y ′ ( x ) = 9 − 5 sec 2 ( x ) and the second derivative is y ′′ ( x ) = − 10 sec 2 ( x ) tan ( x ) . The critical points are at c = ± arccos ( 3 5 ) , with the function being concave upward on ( − 2 π , 0 ) and concave downward on ( 0 , 2 π ) .
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