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It should be 1 and 3 ?? please correct me if I'm wrong.

yes, you are correct.
why you think about 3rd. i think 1st only
because $\frac{dy}{dx}<0$ for x<3
yes i is only correct keep x=1 and x=-1, 3rd is wrong.
Nitish can you point out how to check increasing or decreasing.
@anu isn't the counter example correct, which i mentioned?
anyway, for negative values which are obviously less than 3, this function will keep on increasing.
1 and 3 both are correct.
@hacker16 how? did you consider my comments?

see this,

f'(x) < 0 on (-∞, 3), thus f(x) is decreasing

f'(x) > 0 on (3, ∞), thus f(x) is increasing
@manu check nitish comment it will clarify everything.
@Anu sir,

if $\frac{dy}{dx}$ > 0 then function is increasing

if  $\frac{dy}{dx}$ < 0 then function is decreasing

if  $\frac{dy}{dx}$ = 0 then function is constant

yep, my reasoning was incorrect.
A function f(x) is said to be strictly decreasing on an interval I if f(b)<f(a) for all b>a.
function for larger values,  gives the smaller result.
I am not so good in maths specially with these graphs.

But if we find pointwise

f(5)=25-30+666=661

f(4)=658

-----------------------------------------------------------

Now again f(2)=658

f(1)=661

f(0)=666

f(-1)=673

So, cannot we tell , it is only increasing function?

Moreover, we get slope, where there is two values x and y

But we are here calculating only one value of x.

Where am I mistaking?

@srestha f(2) = 658 f(1)=661 f(0)=666 f(-1)=673

see here on larger value of x, function f(x) is giving smaller value. see the definition of decreasing function in the above comment by me.
@srestha you need to start checking from left side of the number system while checking for increasing or decreasing,

here also from negative infinity to +3, it is definitely decreasing.

also in f(b)-f(a)/b-a ,put the b and a appropriately and you will get the right answer
$f(x) = x^2 - 6x + 666$

$f'(x) = 2x-6$

1) Find critical points, here $3$ is critical point.

2) find sign in the intervals $(-\infty,3),(3, \infty)$

3) take any values say x=1 in $(-\infty,3)$  then $f'(x)$ is -ve

function is decreasing

4) take any values say x=4 in $(3,\infty)$  then $f'(x)$ is +ve

function is increasing.
yes a) and c) both correct
Yes, I think both a and c are correct.

it will be easy by solving through wavey curve  a,c are the correct options

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