Question

How do I solve this?

Evaluate for h(1/3). Show your work.
h(x) = 2x2 - x

Answer 1

Simply substitute in the given value (1/3) in place of x in the function.

So 2x^2 - x -> 2(1/3)^2 - (1/3)

Answer 2

ohh

Answer 3

substitute 1/3 in place of x

h(1/3) = 2(1/3)^2 - (1/3) = 2 (1/9) - (1/3) = (2/9) - (1/3) = (2-3)/9 = -1/3

Answer 4

I think it should be -1/9 there at the end Harkirat ;)

Answer 5

h(x) = 2x^2 - x
= 2(1/3)^2 - 1/3
= 0.4356 - .33
= 0.1056

is that right?

Answer 6

yeah, u r right, i made a typo....☺

Answer 7

Oh, I didn't see your explanation Harkirat :P

Answer 8

substitute 1/3 in place of x

h(1/3) = 2(1/3)^2 - (1/3) = 2 (1/9) - (1/3) = (2/9) - (1/3) = (2-3)/9 = -1/9

Answer 9

More or less LifeIsADangerousGame, by try and use rational expressions where you can to preserve accuracy. Every time you round it increases the error, and adds to the accumulative error in the end.

Answer 10

Ration expressions like how?

Answer 11

Like rather than using .3333, use 1/3. Learn the rules for fractions and for exponents and apply them when solving.

Answer 12

ohh ok

Answer 13

Just like Harkirat?

Answer 14

Yes just like Harkirat. Also I'm not sure where those numbers came from, but they don't appear to be correct (your decimal numbers).

Answer 15

Well I did 1/3 and my calculate got me .333333333 so I rounded it to .33

Answer 16

I can see that but the 0.4356?

Answer 17

and then I did .33 times 2, then .66 X .66 and then got .4356

Answer 18

Ah I see. That's not quite right.

x(y)^2 = x * (y^2) not (x*y)^2.

Answer 19

ohhh ok, I see