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If 
f(x)=(-5)/(8x^(3)), what is the value of 
f(-1), to the nearest ten-thousandth (if necessary)?
Answer:

If f(x)=58x3 f(x)=\frac{-5}{8 x^{3}} , what is the value of f(1) f(-1) , to the nearest ten-thousandth (if necessary)?\newlineAnswer:

Full solution

Q. If f(x)=58x3 f(x)=\frac{-5}{8 x^{3}} , what is the value of f(1) f(-1) , to the nearest ten-thousandth (if necessary)?\newlineAnswer:
  1. Substitute xx with 1-1: To find the value of f(1)f(-1), we need to substitute xx with 1-1 in the function f(x)=58x3f(x) = \frac{-5}{8x^{3}}.\newlineCalculation: f(1)=58(1)3=58(1)=58=58f(-1) = \frac{-5}{8(-1)^{3}} = \frac{-5}{8(-1)} = \frac{-5}{-8} = \frac{5}{8}.
  2. Calculate f(1)f(-1): Now, we need to express the result as a decimal rounded to the nearest ten-thousandth.\newlineCalculation: 58\frac{5}{8} as a decimal is 0.6250.625.

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