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3 years ago

Find the range of values for whichx2–5x + 6 < 0

Mathematics

1 answer:

meriva3 years ago

7 0

Answer:

-2<x<3

or x∈(-2,3)

Step-by-step explanation:

x²-5x<-6

add (-5/2)² i.e.,25/4

x²-5x+25/4<-6+25/4

(x-5/2)²<(-24+25)/4

(x-5/2)²<1/4

(x-5/2)²<(1/2)²

|x-5/2|<1/2

-1/2<x-5/2<1/2

Add 5/2

-1/2+5/2<x-5/2+5/2<1/2+5/2

-2<x<3

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

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Answer:

Step-by-step explanation:

Given the function f(x) = 8x³ − 12x² − 48x, <em>the critical point of the function occurs at its turning point i,e at f'(x) = 0</em>

First we have to differentiate the function as shown;

$f'(x)= 3(8)x^{3-1}- 2(12)x^{2-1} - 48x^{1-1}\\ \\f'(x) = 24x^2 - 24x-48x^0\\\\f'(x) = 24x^2 - 24x-48\\\\At \ the\turning\ point\ f'(x)= 0\\24x^2 - 24x-48 = 0\\\\\\$

$Dividing \ through \ by \ 24\\\\x^2-x-2 = 0\\\\On \ factorizing\\\\x^2-2x+x-2 = 0\\\\x(x-2)+1(x-2) = 0\\\\(x-2)(x+1) = 0\\\\x-2 = 0 \ and \ x+1 = 0\\\\x = 2 \ and \ -1$

Hence the critical numbers of the function are (2, -1)

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3 years ago

Find the range of values for whichx2–5x + 6 < 0​

Find the range of values for whichx2–5x + 6 < 0