Consider the following function.
[tex]f(x)=6 \sin (x)+\sin (2 x), \quad[0,2 \pi][/tex]
Find [tex]f^{\prime}(x)[/tex].
[tex]f(x)=6 \cos (x)+2 \cos (2 x)[/tex]
Find [tex]f^{\prime \prime}(x)[/tex].
[tex]f^{\prime}(x)=-6 \sin (x)-4 \sin (2 x)[/tex]
Find the points of inflection of the graph of the function. (Round your answers to three decimal places. If an answer does not exist, enter DNE.)
smallest x-value (x, y) = (
(x, y) = ([tex]\pi[/tex], 0,)
largest x-value (x, y) = (
concave upward ([2.418, [tex]\pi[/tex]) \cup (3.865,2 [tex]\pi[/tex])[/tex])
concave downward ([0,2.418) \cup ([tex]\pi[/tex], 3.865)[/tex])
Answer (1)
Find the second derivative f ′′ ( x ) = − 6 sin ( x ) − 4 sin ( 2 x ) .
Solve f ′′ ( x ) = 0 to find potential inflection points: x = 0 , π , 2 π , 2.419 , 3.864 .
Check the concavity change at each point to confirm inflection points.
The inflection points are ( 0 , 0 ) , ( π , 0 ) , ( 2 π , 0 ) , ( 2.419 , 3.612 ) , and ( 3.864 , − 3.612 ) .
( 0 , 0 ) ( 2 π , 0 )
Explanation
Find the derivatives We are given the function f ( x ) = 6 sin ( x ) + sin ( 2 x ) on the interval [ 0 , 2 π ] . We need to find the inflection points of the graph of the function. Inflection points occur where the second derivative changes sign. First, we find the first and second derivatives.
State the derivatives The first derivative is given as f ′ ( x ) = 6 cos ( x ) + 2 cos ( 2 x ) .
The second derivative is given as f ′′ ( x ) = − 6 sin ( x ) − 4 sin ( 2 x ) .
Set the second derivative to zero To find the inflection points, we need to solve f ′′ ( x ) = 0 for x in the interval [ 0 , 2 π ] .
− 6 sin ( x ) − 4 sin ( 2 x ) = 0 We can use the double angle identity sin ( 2 x ) = 2 sin ( x ) cos ( x ) to rewrite the equation as: − 6 sin ( x ) − 4 ( 2 sin ( x ) cos ( x )) = 0 − 6 sin ( x ) − 8 sin ( x ) cos ( x ) = 0 Factor out − 2 sin ( x ) :
− 2 sin ( x ) ( 3 + 4 cos ( x )) = 0
Solve for x This equation is satisfied if sin ( x ) = 0 or 3 + 4 cos ( x ) = 0 .
If sin ( x ) = 0 , then x = 0 , π , 2 π in the interval [ 0 , 2 π ] .
If 3 + 4 cos ( x ) = 0 , then cos ( x ) = − 4 3 . The solutions for x in the interval [ 0 , 2 π ] are x = arccos ( − 4 3 ) and x = 2 π − arccos ( − 4 3 ) .
Approximate the solutions Using a calculator, we find that arccos ( − 4 3 ) ≈ 2.4188 and 2 π − arccos ( − 4 3 ) ≈ 3.8642 . Rounding to three decimal places, we have x ≈ 2.419 and x ≈ 3.864 .
Check concavity Now we need to check if the concavity changes at these points. We have the potential inflection points x = 0 , π , 2 π , 2.419 , 3.864 .
We can analyze the sign of f ′′ ( x ) in the intervals ( 0 , 2.419 ) , ( 2.419 , π ) , ( π , 3.864 ) , and ( 3.864 , 2 π ) .
For x ∈ ( 0 , 2.419 ) , let's take x = 1 . Then f ′′ ( 1 ) = − 6 sin ( 1 ) − 4 sin ( 2 ) ≈ − 6 ( 0.841 ) − 4 ( 0.909 ) ≈ − 5.046 − 3.636 = − 8.682 < 0 , so the function is concave down.
For x ∈ ( 2.419 , π ) , let's take x = 3 . Then 0"> f ′′ ( 3 ) = − 6 sin ( 3 ) − 4 sin ( 6 ) ≈ − 6 ( 0.141 ) − 4 ( − 0.279 ) ≈ − 0.846 + 1.116 = 0.27 > 0 , so the function is concave up.
For x ∈ ( π , 3.864 ) , let's take x = 3.5 . Then f ′′ ( 3.5 ) = − 6 sin ( 3.5 ) − 4 sin ( 7 ) ≈ − 6 ( − 0.351 ) − 4 ( 0.657 ) ≈ 2.106 − 2.628 = − 0.522 < 0 , so the function is concave down.
For x ∈ ( 3.864 , 2 π ) , let's take x = 6 . Then 0"> f ′′ ( 6 ) = − 6 sin ( 6 ) − 4 sin ( 12 ) ≈ − 6 ( − 0.279 ) − 4 ( − 0.536 ) ≈ 1.674 + 2.144 = 3.818 > 0 , so the function is concave up.
Find the y-values Since the concavity changes at x = 0 , π , 2 π , 2.419 , 3.864 , these are indeed inflection points. Now we find the corresponding y -values:
f ( 0 ) = 6 sin ( 0 ) + sin ( 0 ) = 0
f ( π ) = 6 sin ( π ) + sin ( 2 π ) = 0
f ( 2 π ) = 6 sin ( 2 π ) + sin ( 4 π ) = 0
f ( 2.419 ) = 6 sin ( 2.419 ) + sin ( 2 ( 2.419 )) ≈ 6 ( 0.679 ) + ( − 0.462 ) ≈ 4.074 − 0.462 = 3.612
f ( 3.864 ) = 6 sin ( 3.864 ) + sin ( 2 ( 3.864 )) ≈ 6 ( − 0.679 ) + ( 0.462 ) ≈ − 4.074 + 0.462 = − 3.612 So the inflection points are ( 0 , 0 ) , ( π , 0 ) , ( 2 π , 0 ) , ( 2.419 , 3.612 ) , and ( 3.864 , − 3.612 ) .
List the inflection points The smallest x -value is 0 , so the inflection point is ( 0 , 0 ) .
The next x -value is 2.419 , so the inflection point is ( 2.419 , 3.612 ) .
The next x -value is π , so the inflection point is ( π , 0 ) .
The next x -value is 3.864 , so the inflection point is ( 3.864 , − 3.612 ) .
The largest x -value is 2 π , so the inflection point is ( 2 π , 0 ) .
Determine concavity The intervals where the function is concave upward are ( 2.419 , π ) ∪ ( 3.864 , 2 π ) .
The intervals where the function is concave downward are ( 0 , 2.419 ) ∪ ( π , 3.864 ) .
Examples
Understanding inflection points is crucial in various real-world applications. For instance, in economics, identifying inflection points on a revenue curve can help businesses determine when the rate of revenue growth starts to slow down, signaling a need for strategic adjustments. Similarly, in physics, analyzing the motion of an object, inflection points on a position-time graph indicate moments of maximum or minimum acceleration. These points provide valuable insights for optimizing strategies and predicting future trends.
