Question

the values of x for which the function f(x) =x²-4x+1 is decreasing are
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Answer 1

Given

we have given function = x²-4x+1

To find

we have find value of x at which function is decreasing

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Step by step explanation:

1. First we will find derivative of the given function.
2. Now,we place derivative of the function equals to zero and find x's value.
3. If the value of x is ≽0 then the function is increasing or if x is ≼0 then the function is decreasing.

=>f(x)= x²-4x+1

finding derivatives

=>f'(x)= 2x-4

f'(x)=0

=>2x-4= 0

=>2x= 4

=>x= 4÷2

=>x=2

Our interval breaks in :

-∞-------------2-------------∞

(-∞,2) x< 2 f(x) is strictly decreasing

(2,∞) x> 2 f(x) is strictly increasing

Check:

(-∞,2) put any values between this in the function

put x = 1

f'(x) = 2x-4

f'(1)= 2-4= -2<0 [ strictly decreasing]

(2,∞) put any values between this

put x = 3

f'(x)= 2x-4

f'(3)= 2(3)-4= 2 >0 [strictly increasing]