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Calculating Relative Minimum and Points of Inflection

This text explains how to calculate the relative minimum and points of inflection of a function using derivatives.

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Calculating Relative Minimum and Points of Inflection

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  1. Part (a) f(x) = kx1/2 – ln x f’(x) = ½kx-1/2 – 1/x f’’(x) = -1/4 kx-3/2 + x-2

  2. Part (b) Since x=1 makes the 2nd derivative positive,our critical point is a relative minimum. If k=2, then the2nd derivative becomes: f’(x) = ½kx-1/2 – 1/x Critical points occur wherethe 1st derivative is zero. f’’(x) = -1/4(2)x-3/2 + x-2 ½kx-1/2 – 1/x = 0 f’’(x) = -½x-3/2 + x-2 ½k(1)-1/2 – 1/(1) = 0 Plugging in the criticalpoint x=1 gives us: ½k - 1 = 0 f’’(x) = -½(1)-3/2 + (1)-2 k = 2 f’’(x) = -½ + 1 = ½

  3. 1/4k x = 1 -1/4k x + 1= 0 k x = 4 Part (c) Points of inflection occur when the 2nd derivative is zero. f’’(x) = -1/4 kx-3/2 + x-2 -1/4 kx-3/2 + x-2 = 0 x-2(-1/4 kx1/2 + 1)= 0 (1/x2)(-1/4 kx1/2 + 1)= 0 (Can’t benegative)

  4. f(x) = k x – ln x k x – ln x = 0 k x = ln x (x-axis crossing) k e4 = 4 k x = 4 Part (c) Remember our original function... ln x = 4 eln x = e4 x = e4 k = 4/e2

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