Question

Find the range of the given function.

f(x) = |x-3| + |x²+6|


Restriction : Solve this question without using concept of minima (basically you are not allowed to use differentiation.)

Answer 1

For X>3 , mod will be opened without any sign

x²+6+x-3=x²+x+3 ------------------1.

Completing the square

(x+½)²+11/4

R-[11/4, ♾️)

For X<3

x²+6-x+3= x²-x+9 -------------------2.

completing the square

(x-½)²+35/4

R-[35/4, ♾️)

Taking union of both ranges

Final range R=[11/4, ♾️)

Answer 2

Answer:

Given x>0

⇒3x>0

⇒−3x<0

⇒2−3x<2

⇒f(x)<2

∴ Range of f=(−∞,2)

(ii) Since, for any real number x, x

2

≥0

⇒x

2

+2≥0+2

⇒x

2

+2≥2

⇒f(x)≥2

∴ Range of f=[2,∞)

(iii) f(x)=x,x is a real number

It is clear that the range of f is the set of all real numbers

∴ Range of f=R