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tester [92]
3 years ago
12

Asap: please can you help me with this question

Mathematics
2 answers:
Leya [2.2K]3 years ago
6 0

Answer:

Step-by-step explanation:

1 Factor {x}^{2}-5x+6x

2

−5x+6.

(x-3)(x-2)<0

(x−3)(x−2)<0

2 Solve for xx.

x=3,2

x=3,2

3 From the values of xx above, we have these 3 intervals to test.

\begin{aligned}&x<2\\&2<x<3\\&x>3\end{aligned}

​

 

x<2

2<x<3

x>3

​

4 Pick a test point for each interval.

For the interval x<2x<2:

Let's pick x=0x=0. Then, {0}^{2}-5\times 0+6<00

2

−5×0+6<0.

After simplifying, we get 6<06<0, which is false.

Drop this interval..

For the interval 2<x<32<x<3:

Let's pick x=\frac{5}{2}x=

25

​

. Then, {(\frac{5}{2})}^{2}-5\times \frac{5}{2}+6<0( 25) 2−5× 25+6<0.

After simplifying, we get -0.25<0−0.25<0, which is true.

Keep this interval..

For the interval x>3x>3:

Let's pick x=4x=4. Then, {4}^{2}-5\times 4+6<04 2 −5×4+6<0.

After simplifying, we get 2<02<0, which is false.

Drop this interval..

5 Therefore,

2<x<3

2<x<3

arsen [322]3 years ago
3 0

Answer:

Step-by-step explanation:

x^2 - 5x + 6 = 0

(x - 2)(x - 3)

So the critical point are x = 2 and x = 3.

The graph of the function opens upwards and pass through the x axis at (2, 0) and (3, 0).

So the required range is the values of f(x) between  when x = 2 and x = 3

that is  2 < x < 3.

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