Question

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 1

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. ]

Answer 2

Answer:

b)  [3 , ∞)

Step-by-step explanation:

We have to find the values of x for which f(x) is increasing;

Since, the given function is f(x) = $x^{2} -6 \times x +3$

For an increasing function , we know that $f'(x) \geq 0$

The function is increasing at the pint when $f'(x)\geq 0$

We have to calculate the range of x by differentiating the given function and making it greater than zero.

∴$f'(x) = 2 \times x -6$

∴ Applying, $f'(x) \geq  0$

$2 \times x - 6 \geq  0$

$2 \times x \geq  6$

$x\geq  3$

∴ For every value of $x\geq 3$ , the function f(x) is incresing

∴  x ∈ [3 , ∞)