t(x) = (x + 3)(x - 4)^2 has an inflection point where x = -5/3 .?

Question

t(x) = (x + 3)(x - 4)^2 has an inflection point where x = -5/3 .



True
or
False

Answer 1

False, pt of inflection of it is at x=5/3

Working:

T=(x + 3)(x - 4)(x - 4)
T=(x+3)(x^2-8*x+16)
T=x^3-8*x^2+16*x+3*x^2-24*x+48
T=x^3-5*x^2-8*x+48
dT/dx = 3x^2-10x-8
d2T/dx2= 6x-10

point of inflection: d2T/dx2 = 0

therefore

6x-10 = 0
6x=10
x=5/3

Answer 2

False, simplification is X^3-5x^2-8X+48, first derivative is 3X^2-10X-8, second derivative is 6X-10, set that to zero, you get X=5/3 as a possible inflection point, after checking that the second derivative has different sides before and after x=5/3, it now is confirmed that it is an inflection point

Answer 3

Inflection factors are the place the 2nd by-product = 0 f (x) = x^3 - 3x^2 + 6x f ' (x) = 3x^2 - 6x f ' ' (x) = 6x - 6 6x - 6 = 0 x = a million <=== inflection factor y = a million^3 - 3 * a million^2 + 6 * a million = 4 (a million , 4) <=== inflection factor

Answer 4

Product rule:-
t `(x) = 1.(x - 4)² + 2.(x - 4) .(x + 3)
t `(x) = (x - 4)² + 2x² - 2x - 24
t "(x) = 2.(x - 4) + 4x - 2
t "(x) = 6x - 10 = 0 for point of inflection.
6x = 10
x = 5 / 3
Question states - 5 / 3
FALSE

Answer 5

points of inflection occur at the second derivative

t'(x)=1(x-4)^2+(x+3)2(x-4)(1) = (x-4)^2+2x^2-2x-24
t"(x) = 2(x-4)(1)+4x-2 = 6x-10
6x-10=0
x=5/3

False

Answer 6

t(x)=x^3 -5x^2 -8x +48

taking first derivative:

t'(x)=3x^2-10x-8

Second Derivative:
t''(x)=6x-10

Inflection point on t''(x)=0

so 0 = 6x -10

or x= 5/3 (no minus)