Let f(x)=24/1+3e−1.3x .over what interval is the growth rate of the function decreasing? (−∞, ln 31.3)(ln1.33, ∞)(ln31.3, ∞) (−∞, ln 1.33)

  (ln(3)/1.3, ∞)

Step-by-step explanation:

The growth rate of the function is given by the first derivative:

  f'(x) = -93.6e^(-1.3x)/(1 +3e^(-1.3x))^2

The interval on which this is decreasing can be found by looking at the second derivative:

  f''(x) = -93.6(1.3e^(-1.3x)(1+3e^(-1.3x))^-2 -e^(-1.3x)(-2)(3)(-1.3)e^(-1.3x)(1+3e^(-1.3x))^-3)

This will be zero when ...

  -3e^(-1.3x) +1 = 0 . . . . . . the only numerator factor that can be zero

  e^(-1.3x) = 1/3 . . . . . . . . . add 3e^(-1.3x), divide by 3

  -1.3x = ln(1/3) = -ln(3) . . take the logarithm

  x = ln(3)/1.3 ≈ 0.845086 . . . . . . . divide by the coefficient of x

The function f(x) is decreasing for all values of x larger than this.

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step-by-step explanation:

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step-by-step explanation:

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