Graphing question - Calculus

Question

produce the graph of f : f(x)= 1+(1/x) +(4/x^2) +(1/x^3)

Use calculus to find the intervals of increase and decrease and intervals of concavity(up and down).

Answer 1

f(x)= 1+(1/x) +(4/x^2) +(1/x^3)

A more convenient form is:

f(x) = 1 + x^-1 + 4x^-2 + x^-3

To get maxima and minima, take the 1st derivative and set = to zero:

f'(x) = -x^-2 -8x^-3 -3x^-4 = 0

To solve, multiply both sides by -x⁴ :

f'(x) = x² + 8x + 3 = 0

Solve by completing the square:

x² + 8x + 16 = 13

(x + 4)² = 13

x + 4 = ±√13, x = -4 ± 3.61 = -0.39, -7.61

for other features, try the 2nd derivative:

f"(x) = 2x + 8 = 0, so x = -4

Start the plotting process, by finding a few easy points:

f(0) = Undefined---ie. +/- infinity

Note also that the function goes to +1 for large + or - x

f(1) = 7

f(-1) = 3

f(-0.5) = -8

Continue calculating key points until the shape becomes clear