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Calculus/f(x)= sinx + 1 on [-2pi, 2pi]

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Question
(a) Find the x- and y-intercepts of the function.
(b) Draw a number line for f:
(c) Find all critical numbers for f0:
(d) Draw a number line for f0:
(e) Find all zeros of f00:
(f) Draw a number line for f00:
(g) Find all in‡ection points of f:
(h) Sketch the graph of f. Label all in‡ection points. Label your axes.

Answer
(a) Note that the minimum of a sin() curve is -1 at x=π ± 2nπ for n any integer.
For the curve, they are -π and 3π.  These are all of the x-intercepts. The y-intercept is at x=0.

(b) For a number line, I would be down the values -2π, -3π/2, -π, -π/2, 0, π/2, π, 3π/2, and 2π
as the x values and find y.  Note that these value are all integers.

(c) The critical numbers are the ones given in (a) and ±2nπ for all integers n,
for these are the values where the maximum or minimum occurs.
For the curve given, they are at -3π/2, -π/2, π/2, and 3π/2.

(d) What is meant by, 'Draw a number line for f0?'

(e)&(f) What is f00?

(g) Is , "in‡ection" suppose to be intersection?
Those are the points where it crosses the x-axis.

(h) It would be a curve that starts at y=1 at x=-2π, rise and levels off to y=2 and x=-3π/2, decreases to y=1 at x=-π with a downward slope, keeps decreasing and leveling off at y=0 where
x=-π/2, and goes back to y=1 at x=0.  It then repeats the same pattern that was on x in (-2π,0) for the interval (0,2π).

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