Answer:
(a) increasing: (-ln(2)/3, ∞); decreasing: (-∞, -ln(2)/3)
(b) minimum: (-ln(2)/3, (9/8)∛2) ≈ (-0.21305, 1.41741); maximum: DNE
(c) inflection point: DNE; concave up: (-∞, ∞); concave down: DNE
Step-by-step explanation:
The first derivative of f(x) = e^(8x) +e^(-x) is ...
f'(x) = 8e^(8x) -e^(-x)
This is zero at the function minimum, where ...
8e^(8x) -e^(-x) = 0
8e^(9x) -1 = 0 . . . . . . multiply by e^x
e^(9x) = 1/8 . . . . . . . add 1, divide by 8
9x = ln(2^-3) . . . . . . take the natural log
x.min = (-3/9)ln(2) = -ln(2)/3 . . . divide by the coefficient of x, simplify
This value of x is the location of the minimum.
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The function value there is ...
f(-ln(2)/3) = e^(8(-ln(2)/3)) + e^(-(-ln(2)/3))
= 2^(-8/3) +2^(1/3) = 2^(1/3)(2^-3 +1)
f(x.min) = (9/8)2^(1/3) . . . . . minimum value of the function
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A graph shows the first derivative to have positive slope everywhere, so the curve is always concave upward. There is no point of inflection. The minimum point found above is the place where the function transitions from decreasing to increasing.
