Question

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = e8x + e−x (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection point. (x, y) = Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)

Answer 1

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.