point of inflection | ab calculus bc,12th grade math

# Point of Inflection

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Point of Inflection is a point at which a curve is changing concave upward to concave downward or vice-versa.
A curve y = f(x) has one of its points x = c as an inflection point
1) if f "(c)= 0 or is not defined and
2) if f "(x) changes sign as x increases through x = c.
The 2nd condition may be replaced by f "'(c ) $\neq$ 0 when f "'(c ) exists.
Thus, x = c is a point of inflection if f "(c) = 0 and F "'(c) $\neq$ 0

## Examples on Point of Inflection

Example 1 : Determine the point of inflection of f(x) = $x^{3}-6x^{2}+11$
Solution : f(x) = $x^{3}-6x^{2}+11$
f '(x) = $3x^{2}-12x$
f "(x) = 6x -12 To find the inflection point set f "(x) = 0
6x -12 = 0
6x = 12
x = 2
To find the y-coordinate of inflection point plug in x =2 in the given equation
f(x) = $x^{3}-6x^{2}+11$
f(2) = $2^{3}-6(2)^{2}+11$
= 8 - 24 + 11
f(2) = -5
Inflection point (2,-5).

Example 2 : Determine the point of inflection of f(x) = $x^{4}-6x^{2}+4$
Solution : f(x) = $x^{4}-6x^{2}+4$
f '(x) = $4x^{3}-12x$
f "(x) = $12x^{2}-12$
To find the inflection point set f "(x) = 0
$12x^{2}-12$ = 0
12($x^{2}$ -1)= 0
$x^{2}$ -1 = 0
$x^{2}$ = 1
x = $\pm$ 1
To find the y-coordinate of inflection point plug in x =1 and x = -1 in the given equation
f(x) = $x^{4}-6x^{2}+4 f(1) =$(1)^{4}-6(1)^{2}+4
= 1 - 6 + 4
f(1) = -1
f(x) = $x^{4}-6x^{2}+4 f(-1) =$(-1)^{4}-6(-1)^{2}+4
= 1 - 6 + 4
f(1) = -1
Inflection points (1,-1) and (-1,-1).

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