If g(x)=x^2/3, show that g'(0) does not exist
and..
if a is not equal to 0, find g'(a).
2006-06-21 15:14:43 · 10 answers · asked by johnnyboy16978 1 in Science & Mathematics ➔ Mathematics
Graph the function g(x) = x^(2/3). The graph has a cusp at zero. The left-hand and right-hand limits ( in the definition of the derivative ) are different, so that g(x) is not differentiable at zero.
The standard power rule will work for g'(a) away from 0.
g'(a) = (2/3) a^(-1/3). ( You can also see that plugging in a=0 causes a division by zero problem! )
2006-06-21 15:24:34 · answer #1 · answered by AnyMouse 3 · 0⤊ 0⤋
g'(0) = 2/3 x ^ -1/3
= 2/3 ( 0 ) ^ -1/3
= 0
2006-06-21 22:21:14 · answer #2 · answered by Navarro Wee 3 · 0⤊ 0⤋
(I'm assuming you mean x^(2/3). If you meant (x^2)/3, then g'(0) *does* exist.)
When you take the derivative of g(x), using the Power Rule, you'll notice that the exponent on the variable is a negative number. As you know, x^-n = 1/(x^n), and you're not allowed to have zero in the denominator of a fraction.
Mind you, I'm not going to work out the 2nd part, because you need to find g'(x) yourself. Hope that helps, though.
EDITED TO ADD: oh well, a couple other people already worked out g'(x) for you instead. Ain't life grand? ;-)
2006-06-21 22:21:27 · answer #3 · answered by Jay H 5 · 0⤊ 0⤋
differentiate
will becomes
g'(x)=2/(3(x^1/3))
if g(0) the the value is infinite thus
a must not equal to zero
g'(a)=2/(3(a^1/3))
where a is not equal to zero.
2006-06-21 22:36:30 · answer #4 · answered by Anonymous · 0⤊ 0⤋
g'(a)=(2/3)a^(-1/3), so when a=0 you get a 0^(-1/3) term. This is 1/(0^1/3) which is 1/0 which is undefined.
2006-06-21 22:21:23 · answer #5 · answered by zee_prime 6 · 0⤊ 0⤋
If I remember correctly, taking the derivative would put x in the denominator raised to the 1/3 power and if x = 0 then g'(0) d.n.e.
2006-06-21 22:24:04 · answer #6 · answered by miss_tex 1 · 0⤊ 0⤋
when we make derevative
g'(x) = (2/3)* x^(-1/3)
= (2) / (3 * cubicroot(x) )
sub x by 0, we get: (2) / (3 * 0) = (2/0) "undefined value"
so g'(0) doesn't exist..
and by sub. by "a":
g'(a) = (2) / (3 * cubicroot(a) )
2006-06-22 03:09:23 · answer #7 · answered by Anonymous · 0⤊ 0⤋
g'(x) = (2/3)* x^(-1/3)
so g'(0) is infinite....doesnot exist
g'(a) = (2/3)* a^(-1/3)
2006-06-21 22:19:19 · answer #8 · answered by sunil 3 · 0⤊ 0⤋
Calculus is the debil eeehhhhh
having bad high school flash backs... gotta hide gotta hide :((
2006-06-21 22:34:20 · answer #9 · answered by Jembee1720 4 · 0⤊ 0⤋
i don't remember calculus. just wondering, if you aren't in engineering, what would you use it for, once you graduate college?
2006-06-21 22:21:19 · answer #10 · answered by miss mary mac 2 · 0⤊ 0⤋