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Differential Calculus

Scorching goIITian

Joined:	2 Mar 2007
Post:	273

11 Sep 2007 00:25:47 IST

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320

app of derivatives......plz

None

if h(x) = 3 f(x²/3) + f(3- x²) where x blong 2 (-3,4)
nd f"(x)>0 where x blong 2 (-3,4)
find intervals og increase nd decrease of h(x)....
........
ans... h(x)is decreasibg in (-3/2, 3/2) nd increasing in (-3,-3/2) U (3/2,4)

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Comments (2)

Ramya Hegde

Blazing goIITian

Joined: 8 Feb 2007

Posts: 612

11 Sep 2007 03:50:16 IST

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Well, i'm getting a slightly different answer. So, i'm posting my method, do tell me where i'm wrong.

My answer comes out be

h(x) is increasing from (-3/2, 0) (3/2,4)

and decreasing in (-3/-3/2) (0,3/2)

I've done it as below!!

Given that,

f^''(x)>0, hence f^'(x) is increasing,

i.e., for any a<b, f^'(a)<f ^'(b)

Now,

h(x) will be increasing or decreasing if h^'(x) is +ve or -ve.

h^'(x) = 3.f^'(x²/3).(2x/3) + f^'(3-x²).(-2x)

= 2x [ f^'(x²/3) - f^'(3-x²)]

Now, h^'(x) is increasing, if [ f^'(x²/3) - f^'(3-x²)] is +ve.

Hence, we should find the intervals in which x²/3>3-x²

for, x²/3 >3-x²

x²>9/4

i,e, x>3/2 or x<-3/2

Hence for this interval,[ f^'(x²/3) - f^'(3-x²)] is positive.

But, between -3 and -3/2, x is negative, so the whole expression becomes -ve.

Hence h(x) is decreasing. and for (3/2, 4), the expression is positive as x is also positive.

Similarly for the rest of the interval, it is showing the increasing and decreasing trend.

Hence, my answer is

h(x) is increasing from (-3/2, 0) (3/2,4)

and decreasing in (-3/-3/2) (0,3/2)

U'll get the answer mentioned if u ignore x.

So plzz do tell me the mistake in my solution.

Thanks!!!

shine

Scorching goIITian

Joined: 2 Mar 2007

Posts: 273

11 Sep 2007 17:31:24 IST

Like 0 people liked this

well ramya.......i 2 have adopted d same approach nd getting d same
ans
however
in the solution given x=0 is not taken as a critical point nd hence
d difference in ans arises
lets see wat others say

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