What is the largest interval on which the function is concave down for f(x)=5sin⁡(x)+(sin⁡(x))2 over the interval [−π3,2π3]?

Question

enimane69r · Accepted Answer

Step 1(0,2π3], open on the left and closed on the right.Explanation:f(x) is periodic with period 2π.The half of a full wave, in (0,π), is up, up to 8 units at x=π2, while the shorter half, in (π,2π) goes down up to -4 units at x=3π2.Step 2The tangent crosses the wave at x=0,π,2π, to turn the curve from concavity to convexity, and vice versa. Of course, viewing from the x-axis, it is looking concave, both sides.

conduchafr4 · Accepted Answer

Step 1f′(x)=5cos⁡x+2sin⁡xcos⁡x=5cos⁡x+sin⁡(2x)f″(x)=−5sin⁡x+2cos⁡(2x)=−5sin⁡x+1(1−2sin2x)f″(x)=−4sin2x−5sin⁡x+2Solving fx)=0 we find the partition numbers usingsin⁡x=−5+578The only value of x in the interval with this sine is arcsin⁡(−5+578)≈0.32Step 2For −π3&amp;lt;x&amp;lt;0.3244, we have fx)&amp;gt;0, so f is concave up. (Use x=0 as a test number.)For 0.3244&amp;lt;x&amp;lt;2π3, we have fx)&amp;lt;0, so f is concave down. (Use x=π2 as a test number.)So the largest interval is the closed interval [arcsin⁡(−5+578),2π3]Which is about [0.3244,2π3]

What is the largest interval on which the

Answered question

Answer & Explanation

New Questions in Calculus 1