Angle Abc and bca are supplimentary. Angle abc is 140 degrees. If angle bca is named as x over 2 plus ten, what could be the val
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Answer:
- increasing: (π/2, 3π/2)
- decreasing: [0, π/2) ∪ (3π/2, 2π]
- minimum: -16 at x=π/2
- maximum: 16 at x=3π/2
Step-by-step explanation:
If all you want are answers to the questions, a graphing calculator can provide them quickly and easily. (see attached)
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If you need an algebraic solution, you need to find the zeros of the derivative.
f'(x) = -16cos(x)sin(x) -16cos(x) = -16cos(x)(sin(x) +1)
The product is zero where the factors are zero, at x=π/2 and x=3π/2.
These are the turning points, where the function changes from decreasing to increasing and vice versa.
(sin(x)+1) is non-negative everywhere, so the sign of the derivative is the opposite of the sign of the cosine function. This tells us the function f(x) is increasing on the interval (π/2, 3π/2), and decreasing elsewhere (except where the derivative is zero).
The function local extrema will be where the derivative is zero, so at f(π/2) (minimum) and f(3π/2) (maximum). We already know that cos(x) is zero there, so the extremes match those of -16sin(x).
