IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Taara

16) we know this thm but how to Prove it?

f be function on [a , b] such that f'(x) > 0 all xE (a, b)
then f is increasing function on (a,b)

only one or no root ...is this enough?

sandeepramesh

what do you wanna prove?

arpan1

ya, pls explain clearly

ramkumar_november

i guess this is the answer

$lim_{h\longrightarrow 0} \frac{f(b)-f(a)}{b-a}\;>0$

$so\;\;f(b)>f(a)$

now we have b>a => f(b)>f(a)

so f(x) is an increasing function in [a,b]

dheer_07

TARA U CAN DO IT WITHH LMVT

USING LMVT ONCE U PROVE F[B] >.F[A]

NOW AGAIN USE LMVT IN[B,C]

N PROVE F[C]>F[B]

N SO ON HENCE FX IS INC FUNC

HOPE U GOT IT

PLZ RATE ME USEFUL!!

CHEEERSSSS!!!!

chemicaltester

sorry ramkumar
u cannot say by ur equation that f(b) >f(a)
by taht relation
this is true only if b>a

if bthen to the derivative comes positive

computer001

i think it is obvious from the fact tht the function has positive slope(i.e f'(x)>0

x

[a,b]..so as x inc f(x) increases

dheer_07

HEY IVE DONE, WATS PROB THEN

vacuumhead_pratyush

here, r v supposed 2 prove that if the slope is +ve , how is the function increasing ???????????

vacuumhead_pratyush

here, r v supposed 2 prove that if the slope is +ve , how is the function increasing ???????????

ITZ OBVI.

dheer_07

CUMMON YAAR ITS NCERT QUES

MANY OBVIOUS QUESTIONS R THERE

HAHAHAHA!!!!!

Chandravadan

Hey guys, choose a perfectly increasing fn like a linear fn, go on checking till you are satisfied!

vacuumhead_pratyush

Taara

ya its very obvious like i said.
but they require proof for it...thats why i asked.

thanks.

RyuAmakusa

f'(x) = Lt [f(x) - f(a)] / [x-a]
x->a

if f'(x) > 0 then f(x)>f(a) if x>a and vice versa from the above definition
=>f(x) is increasing.

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