Consider the following function:
[tex]f(x) = 6\cos(x) + 2\cos(2x), \quad (0, 2\pi)[/tex]
Find [tex]f'(x)[/tex].
[tex]f'(x) = -6\sin(x) - 4\sin(2x)[/tex]
Find [tex]f''(x)[/tex].
Find the points of inflection of the graph of the function. (Round your answers to three decimal places. If a value does not exist, enter DNE.)
(x, y) = ( [ ] , [ ] )
(x, y) = ( [ ] , [ ] )
(x, y) = ( [ ] , [ ] )
(x, y) = ( [ ] , [ ] ) largest x-value
concave upwards [ ]
concave downwards [ ]
Answer (2)
Find the second derivative f ′′ ( x ) = − 6 sin ( x ) − 4 sin ( 2 x ) .
Set f ′′ ( x ) = 0 and solve for x , obtaining x = π , x ≈ 2.419 , and x ≈ 3.864 .
Determine the concavity of f ( x ) on the intervals ( 0 , 2.419 ) , ( 2.419 , π ) , ( π , 3.864 ) , and ( 3.864 , 2 π ) .
The inflection points are approximately ( 2.419 , − 4.5 ) , ( π , − 4 ) , and ( 3.864 , − 4.5 ) , with concavity changing at these points. The intervals of concavity are: concave downward on ( 0 , 2.419 ) and concave upward on ( 2.419 , π ) ∪ ( π , 3.864 ) ∪ ( 3.864 , 2 π ) .
The x values of the inflection points are: 2.419 , π , 3.864 .
Explanation
Problem Setup We are given the function f ( x ) = 6 cos ( x ) + 2 cos ( 2 x ) and its second derivative f ′′ ( x ) = − 6 sin ( x ) − 4 sin ( 2 x ) . We need to find the inflection points of the graph of the function on the interval ( 0 , 2 π ) . Inflection points occur where the second derivative changes sign, which typically happens when f ′′ ( x ) = 0 .
Finding Potential Inflection Points First, we set f ′′ ( x ) = 0 and solve for x :
− 6 sin ( x ) − 4 sin ( 2 x ) = 0 Using the identity sin ( 2 x ) = 2 sin ( x ) cos ( x ) , we can rewrite the equation as: − 6 sin ( x ) − 8 sin ( x ) cos ( x ) = 0 Factoring out − 2 sin ( x ) , we get: − 2 sin ( x ) ( 3 + 4 cos ( x )) = 0
Solving for x This equation is satisfied if either sin ( x ) = 0 or 3 + 4 cos ( x ) = 0 .
For sin ( x ) = 0 , the solution in the interval ( 0 , 2 π ) is x = π .
For 3 + 4 cos ( x ) = 0 , we have cos ( x ) = − 4 3 . Let's find the values of x in the interval ( 0 , 2 π ) that satisfy this equation. We have x 1 = arccos ( − 4 3 ) and x 2 = 2 π − arccos ( − 4 3 ) . Using a calculator, we find that x 1 ≈ 2.419 and x 2 ≈ 3.864 .
Determining Concavity Now we need to determine the intervals where 0"> f ′′ ( x ) > 0 (concave up) and f ′′ ( x ) < 0 (concave down) by testing values in the intervals ( 0 , 2.419 ) , ( 2.419 , π ) , ( π , 3.864 ) , and ( 3.864 , 2 π ) .
Let's test x = 1 in ( 0 , 2.419 ) : 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 downward in this interval. Let's test x = 3 in ( 2.419 , π ) : 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 upward in this interval. Let's test x = 4 in ( π , 3.864 ) : 0"> f ′′ ( 4 ) = − 6 sin ( 4 ) − 4 sin ( 8 ) ≈ − 6 ( − 0.757 ) − 4 ( 0.989 ) ≈ 4.542 − 3.956 = 0.586 > 0 , so the function is concave upward in this interval. Let's test x = 5 in ( 3.864 , 2 π ) : 0"> f ′′ ( 5 ) = − 6 sin ( 5 ) − 4 sin ( 10 ) ≈ − 6 ( − 0.959 ) − 4 ( − 0.544 ) ≈ 5.754 + 2.176 = 7.93 > 0 , so the function is concave upward in this interval.
Finding y-coordinates Since the concavity changes at x = 2.419 , x = π , and x = 3.864 , these are the x -coordinates of the inflection points. Now we need to find the corresponding y -coordinates. f ( π ) = 6 cos ( π ) + 2 cos ( 2 π ) = 6 ( − 1 ) + 2 ( 1 ) = − 6 + 2 = − 4 f ( 2.419 ) = 6 cos ( 2.419 ) + 2 cos ( 2 × 2.419 ) ≈ 6 ( − 0.75 ) + 2 cos ( 4.838 ) ≈ − 4.5 + 2 ( − 0.001 ) ≈ − 4.5 f ( 3.864 ) = 6 cos ( 3.864 ) + 2 cos ( 2 × 3.864 ) ≈ 6 ( − 0.75 ) + 2 cos ( 7.728 ) ≈ − 4.5 + 2 ( − 0.002 ) ≈ − 4.5
Inflection Points and Concavity Therefore, the inflection points are approximately ( 2.419 , − 4.5 ) , ( π , − 4 ) , and ( 3.864 , − 4.5 ) . The function is concave downward on ( 0 , 2.419 ) and concave upward on ( 2.419 , π ) , ( π , 3.864 ) , and ( 3.864 , 2 π ) .
Final Answer The inflection points are ( 2.419 , − 4.5 ) , ( π , − 4 ) , and ( 3.864 , − 4.5 ) . The function is concave downward on the interval ( 0 , 2.419 ) and concave upward on the intervals ( 2.419 , π ) , ( π , 3.864 ) , and ( 3.864 , 2 π ) .
Examples
Understanding inflection points and concavity is crucial in various fields. For example, in economics, identifying inflection points on a cost curve can help businesses determine the point at which marginal costs start to increase. In physics, analyzing the concavity of a displacement-time graph can reveal changes in acceleration. These concepts allow professionals to make informed decisions and optimize processes in their respective domains.
The inflection points of the function are approximately at (2.419, -4.5), (π, -4), and (3.864, -4.5). The function is concave downward on the interval (0, 2.419) and concave upward on the intervals (2.419, 2π).
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