Respuesta :
Answer:
the absolute maximum value is 89.96 and
the absolute minimum value is 23.173
Step-by-step explanation:
Here we have cotangent given by the following relation;
[tex]cot \theta =\frac{1 }{tan \theta}[/tex] so that the expression becomes
f(t) = 9t +9/tan(t/2)
Therefore, to look for the point of local extremum, we differentiate, the expression as follows;
f'(t) = [tex]\frac{\mathrm{d} \left (9t +9/tan(t/2) \right )}{\mathrm{d} t}[/tex] = [tex]\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9 \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2}[/tex]
Equating to 0 and solving gives
[tex]\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9 \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2} = 0[/tex]
[tex]t=\frac{4\pi n_1 +\pi }{2} ; t = \frac{4\pi n_2 -\pi }{2}[/tex]
Where n[tex]_i[/tex] is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively
Or we have by chain rule
f'(t) = 9 -(9/2)csc²(t/2)
Equating to zero gives
9 -(9/2)csc²(t/2) = 0
csc²(t/2) = 2
csc(t/2) = ±√2
The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or
in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2
We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;
f(x) for x = π/4, π/2, 3π/2, 7π/2
f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965
f(π/2) = 9·π/2 +9/tan(π/4) = 23.137
f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412
f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96
Therefore the absolute maximum value = 89.96 and
the absolute minimum value = 23.173.
