Suppose that the following group of values was entered into the tvm solver of a graphing calculator: n=240; 1%=13.2; pv=95000; p
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Answer:
answer is attached on the photo
Answer:
Step-by-step explanation:
Given that there is a function of x,
Let us find first and second derivative for f(x)
When f'(x) =0 we have tanx = 1 and hence
a) f'(x) >0 for I and III quadrant
Hence increasing in
and decreasing in
Hence f has a maxima at x = pi/4 and minima at x = 3pi/4
b) Maximum value =
Minimum value =
c)
f"(x) =0 gives tanx =-1
are points of inflection.
concave up in (3pi/4,7pi/4)
and concave down in (0,3pi/4)U(7pi/4,2pi)