If f(x) = |x| + |x - 2| then find the value of F(1.5) for 0 ≤ x ≤ 2
thank you
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Hi,vikram4689 wrote:General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0
f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
f(1.5) = |1.5| + |1.5-2| = 1.5 + |-.5| = 1.5 + .5 =2If f(x) = |x| + |x - 2| then find the value of F(1.5) for 0 ≤ x ≤ 2
Yes, the modulus graph is symmetrical about y-axis but that does not meanvikram4689 wrote:General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
|x| is always non-negative for any value of x.vikram4689 wrote:|x| is +ve for x>0
|x| is -ve for x<0
Hi,vikram4689 wrote:Sorry pals I DO NOT agree on this ( though i made a silly mistake on a -ve sign but concept is correct)
General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
|x| is +ve for x>0
|x| is -ve for x<0
vikram4689 wrote:General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0
f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1