Engineering Board Exam Reviewer · Curve Analysis: Critical Points, Concavity, and Inflection · hard

Find the absolute minimum value of f(x)=x4-4x3+10.

  1. 10
  2. -7
  3. -17
  4. 17

Answer

C. -17

Solution

f'(x)=4x3-12x2=4x2(x-3), so the critical numbers are x=0 and x=3. f' does not change sign at x=0 (no extremum), but it changes from negative to positive at x=3. The minimum is f(3)=81-108+10=-17.

More Curve Analysis: Critical Points, Concavity, and Inflection questions

  1. How many inflection points does f(x)=x5-5x4 have?
  2. Find the maximum value of f(x)=xe-x for x≥0.
  3. For what value of k does f(x)=x3+kx2 have an inflection point at x=2?
  4. Find the absolute maximum value of f(x)=x3-3x on the interval [0,2].
  5. Determine the intervals of concavity of f(x)=x4-6x2.
  6. Find the x-coordinate of the inflection point of f(x)=x3-6x2+9x+1.
  7. Find the local maximum value of f(x)=x3-3x2+4.
  8. On which interval is f(x)=x2-4x increasing?
  9. Find the critical number of f(x)=x2-6x+5.

All Curve Analysis: Critical Points, Concavity, and Inflection questions