Question

the range of values of x in witch f(x)=x2-6x+3 is increasing,is​

Answer 1

Answer:

Function is increasing in the range of values (2,∞)

Step-by-step explanation:

To find the range of values of x for which $f(x)=x^2-6x+3$

Do differentiation of f(x),put the differentiation of f(x) =0,find the value of x.For those values if

f'(x)>0; function is increasing

f'(x)<0; function is decreasing

$f^{'}(x)=2x-6\\\\2x-6=0\\\\2x=6\\\\x=3$

Intervals are

1) (-∞,2)

2) (2,∞)

case1: (-∞,2)

put x= 1

$f^{'}(1)=2(1)-6\\\\=2-6\\\\=-4\\\\=negative$

f(x) is decreasing in the interval (-∞,2)

case2: (2,∞)

put x= 4

$f^{'}(4)=2(4)-6\\\\=8-6\\\\=2\\\\=positive$

f(x) is increasing in the interval (2,∞)

Hope it helps you.

Answer 2

The range of values of x in which f(x) = x² - 6x + 3 is increasing, is(A) [0, 3] B [3, ∞) (C) (-3, 3] (D) (-0, 3]

answer : function is increasing in interval [3, ∞)

explanation : function, f(x) = x² - 6x + 3

differentiating f(x) with respect to x,

df(x)/dx = f'(x) = d(x² - 6x + 3)/dx

⇒f'(x) = 2x - 6

we know, if f'(x) ≥ 0 in (a, b) it means, function is increasing in (a, b).

so, function is increasing only when f'(x) = 2x - 6 > 0

⇒2x > 6

⇒x > 3

hence, for all x ∈ [ 3, ∞), function is increasing.

[note : function is monotonically increasing when f'(x) ≥ 0, for strictly increasing we take f'(x) > 0. here in all options , 3 is included so, we use f'(x) ≥0 also question didn't mention function is strictly increasing or monotonically. ]