CPM Homework Help : APCALC Problem 11-14

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Home > APCALC > Chapter 11 > Lesson 11.1.1 > Problem 11-14

11-14.

Multiple Choice: If $a > 0$ , $f(x) = x + ax^{–2}$ will have a relative minimum at:

$x =\sqrt { 2 a }$

$x =\sqrt [ 3 ] { 2 a }$

$x = a$

$x =\sqrt { 2 }$

$x = ± 2a$

Step 1:

$f^\prime(x)=1-2ax^{-3}=0$

Step 2:

$1=2ax^{-3}$

Step 3:

$x^3=2a$

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