Question

consider the function f(x)=√x-2,g(x)=x+1÷x×x-2x+1​

Answer 1

Answer:

f(x)=loge(√1−x2−x)

√1−x2−x

x∈[−1,1]

x∈[−1,0], √1−x2−x>0

x>0

√1−x2>x⇒1−x2>x2

x∈(0,1√2)

[−1,1√2]

f'(x)=−x√1−x2−1√1−x2−x

f'(x)>0 if −x√1−x2−1>0

or −x−√1−x2>0

or x∈[−1,−1√2)

∴f(x)

[−1,−1√2]

⎛⎜⎝−1√2,1√2

∴ f has local max, at x=−1/√2

limx→1√2loge(√1−x2−x)=−∞

∴ f has no minima.

Answer 2

Answer:

Step-by-step explanation:

Given:

$f(x) = \sqrt{x-2}\\ g(x) = \frac{x+1}{[x^{2} - 2x + 1]}$

Now, For x = 0

$f(0) = \sqrt{0-2}\\ = \sqrt{-2}$  (imaginary)

For x=1

We get,

$f(1) = \sqrt{1-2} \\ = \sqrt{-1}$(imaginary)

For x = 2

$f(2) = \sqrt{2-2} \\ = 0$

Since for the values of below 1, we can see that the results are imaginary then f(x) can be written as,

$f(x) \geq 0\\\\\sqrt{x-2} \geq 0\\\\On squaring both sides, we get,\\x-2 \geq 0\\x\geq 2$

This shows that the domain f f(x0 is {2,∞)

This will act as range for the given equation g(x)