i had a few calc questions if someone can please answer. 1) f(x)arcsec(2/x) ; fi

Question

i had a few calc questions if someone can please answer.

1) f(x)arcsec(2/x) ; find the maximum(x,y) (its not pi/2, or -1/2)

2)A billboardd=86feet wide is perpendicular to a straight road and is 40 feet from the road (see figure). Find the point on the road at which the angle?subtended by the billboard is a maximum. (Round your answer to two decimal places.)

4)And airplane flies at an altitude of 5 miles toward a point directly over an observer. Consider?andxas shown in the figure below.

3)Consider the following.

Explanation / Answer

Relative extrema occur when f'(x) = 0 is f'(x) is undefined. f(x) = arcsec(x) - 9x f'(x) = 1/[|x|*v(x^2 - 1)] - 9 Setting f'(x) = 0, we obtain: 1/[|x|*v(x^2 - 1)] - 9 = 0 1/[|x|*v(x^2 - 1)] = 9 {1/[|x|*v(x^2 - 1)]}^2 = 9^2 1/[x^2*(x^2 - 1)] = 81 x^2*(x^2 - 1) = 1/81 u(u - 1) = 1/81; u = x^2 u^2 - u - 1/81 = 0 u = 1/2 ± v(85)/18 x^2 = 1/2 + v(85)/18 (no negative root since x^2 can't be negative) x = ±v[1/2 + v(85)/18] f'(x) is undefined when it's denominator is 0: |x|*v(x^2 - 1) = 0 x = 0 or x = ±1 At the points x = 0, ±1, ±v[1/2 + v(85)/18], the values of f(x) are: f(0) is undefined f(1) = -9 f(-1) = p + 9 f(v[1/2 + v(85)/18]) ˜ -8.944725876 f(-v[1/2 + v(85)/18]) ˜ 12.08631853 We can see that the maximum value is p + 9 while the minimum value is -9. please rate me

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