Can a local maximum also be a point of inflection? Explain.

1 answer:

7 0

Inflection points are where the function changes concavity. Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. So the second derivative must equal zero to be an inflection point. But don't get excited yet. You have to make sure that the concavity actually changes at that point.

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Find the remaining trigonometric functions of θ based on the given information.csc θ = 257 and cos θ < 0

katrin [286]

Cos(θ) < 0, so we know it would be in Quadrant 2 or 3
then csc(θ) = 257, but csc(θ) = $\frac{1}{sin(θ)}$ = 257
==> sin(θ) = $\frac{1}{257}$
it is positive, so now we can determine that is in Quadrant 2

sin(θ) = opp./hyp both of opp and hyp are positive but adj suppose to negative because that way it leads the cos(θ) < 0
cos(θ) = adj/hyp
Pythereom to find the adj: $257^{2} - 1^{2} = adj ==\ \textgreater \ adj \sqrt{ 257^{2} - 1^{2} } ==\ \textgreater \ adj = 16 \sqrt{258}$

cosθ = $\frac{16 \sqrt{258} }{257} 
$
tanθ = $\frac{1}{16 \sqrt{258} }$
cosθ = $16 \sqrt{258}$