Question

find the domain and range of functions f(x) =xsquare +3x+5 by xsquare - 5x+4​

Answer 1

EXPLANATION.

Domain and Range of the function,

⇒ x² + 3x + 5/x² - 5x + 4.

⇒ f(x) = x² + 3x + 5/x² - 5x + 4.

for domain denominator > 0.

split their domain into middle term split,

⇒ x² - 5x + 4 = 0.

⇒ x² - 4x - x + 4 = 0.

⇒ x ( x - 4 ) -1 ( x - 4 ) = 0.

⇒ ( x - 1 ) ( x - 4 ).

⇒ f(x) = x² + 3x + 5/( x - 1 ) ( x - 4 ) = 0.

⇒ Domain = R - { 1,4}.

To find the Range.

Let the function = y.

f(x) = y.

⇒ x² + 3x + 5/x² - 5x + 4 = y.

⇒ x² + 3x + 5 = y ( x² - 5x + 4 ).

⇒ x² + 3x + 5 = x²y - 5xy + 4y.

⇒ x²y - x² - 5xy - 3x + 4y - 5 = 0.

⇒ x² ( y - 1 ) + x ( -5y - 3 ) + 4y - 5 = 0.

For real roots, D ≥ 0.

⇒ b² - 4ac ≥ 0.

⇒ ( -5Y - 3 )² - 4(y - 1 ) ( 4y - 5 ) ≥ 0.

⇒ 25y² + 9 - 30y - 4 ( 4y² - 5y - 4y + 5 ) ≥ 0.

⇒ 25y² + 9 + 30y - 16y² + 36y - 20 ≥ 0.

⇒ 9y² + 66y - 11 ≥ 0.

to find the value of y = -b ± √b² - 4ac/2a.

⇒ - 66 ± √(66)² - 4(9)(-11) / 18.

⇒ - 11 ± √132/3.

Put it on a wavy curve method,

Range = ( -∞, -11 - √132/3 ] ∪ ( ∞, -11 + √132/3 ]

Answer 2

$\huge\orange{\underline{\underline{\bf{SoluTion:}}}}$

Given,

$\sf{f(x) = \frac{x²+3x+5}{x²-5x+4}}$

$→\;{\sf{\frac{x²+3x+5}{x²-4x-x+4}}}$

$→\;{\sf{\frac{x²+3x+5}{x(x-4)-1(x-4)}}}$

$→\;{\sf{\frac{x²+3x+5}{(x-4)(x-1)}}}$

In real numbers, the denominator can't be equal to Zero (0).

Hence, (x-4)(x-1) ≠ 0

x ≠ 4 or x ≠ 1

So, the domain of the function will be all real numbers except 4 & 1.

$\pink{\underline{\underline{\bf{Required\;AnSweR:}}}}$

The domain ☞ $\fbox\red{\bf{R - {1,4}}}$

Hope It Helps You ✌️