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Unformatted text preview: critical points and the intervals of
Step 4. Find inflection points and the intervals in
which the function is concave up/down. EXAMPLE:
Given a function with the property ekx (2kx 2 − 1) ,
2 f ′( x) = k k > 0 , determine the intervals on which f is increasing and
decreasing and the location of any local
maximum and minimum values of f. SOLUTION:
We find that e kx (2kx 2 − 1)
= 0 when x = ±1 / 2k ,
2 f '( x) = k hence these points are the locations of any local
maximum or minimum values of f. Now we
examine the change of the sign of the derivative
of f at these points. has a relative minimum at p. () If f ′( p ) = 0 and f ′′ p < 0 , then f has a
relative maximum at p. If f ′( p ) = 0 and f ′′( p ) = 0 , then the test fails.
f ′′( p ) > 0 on an interval, then f(x)
is concave up on that interval; if f ′′( p ) < 0 , then Concavity: If x −1/
More free study sheet ′′and practice tests=at: 2k Differentiate both sides: 2x + y + x ⋅ 2 Where is the curve...
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