Question

Help plz

Answer 1

Calculate a and b

Answer 2

Hints:
We make use of the fact that the function is continuous and differentiable.
So the function AND derivative are continuous at x=1.
The two conditions should land you with two equations which should enable you to solve for a and b.

Answer 3

did u get the answer muskan?

Answer 4

They turn out to be nice fractions.

Answer 5

what is the derived \[a + b/ \sqrt{x}\]

Answer 6

We derive with respect to x, so it is
0-b*x^(3/2)/2 = -(b/2)x^(3/2)

Answer 7

Sorry, it should be -(b/2)x^(2/3), or -(b/2)/x^(3/2)

Answer 8

\[- (b/2) * x ^{3/2} \] is it?

Answer 9

Sorry, I got it upside-down, so
\( -(b/2)x^{2/3} \)
would be right

Answer 10

Actually, it is clearer if I wrote
\( -(b/2)x^{-3/2}\)
because this is how we got the derivative

Answer 11

\[\frac{ -b }{ 2} * x ^{-3/2}\] ??

Answer 12

m confused

Answer 13

Take the problem as
differentiate \(a+bx^{-1/2} \)
and use the standard power rule to get
\( 0 -(1/2)bx^{-3/2} \)

Answer 14

and \[1 + a / x-2\]

Answer 15

Use the chain rule:
f(x)=\( 1+ \frac{a}{x-2} \)
so
\( f'(x)= 0 + \frac{a}{(x-2)^2}*1 = \frac{a}{(x-2)^2} \)

Answer 16

ok

Answer 17

\[\lim_{x \rightarrow 1} \frac{ a }{ (x-2)^{2} } = \frac{ a }{ (1-2)^{2} } = \frac{ a }{ 1+4 } = \frac{ a }{ 5 }\]

\[\lim_{a \rightarrow 1} -1/2 * bx ^{-3/2} = -1/2 * b(1)^{-3/2}\]

Is its Correct??

Answer 18

The first limit equals \( \frac{-a}{(1-2)^2} = \frac{-a}{(-1)^2} = -a\)
The second limit is \( \frac{-b/2}{x^{3/2}} = \frac{-b/2}{1^{3/2}}=-\frac{b}{2}\)

Answer 19

why is \[\frac{ -a }{(-1)^{2} }\]

Answer 20

Sorry, my bad.
f'(x) = -a/(x-2)^2 to start with.

Answer 21

ok

Answer 22

So what do you propose for the next step?

Answer 23

continuity

Answer 24

\[\lim_{x \rightarrow 1} 1+\frac{ a }{ 1-2 } = 1 + a/-1 = a=0\]

\[\lim_{x \rightarrow 1 } a+\frac{ b }{ \sqrt{x} } = a+b/1\]

Answer 25

Is it Correct?

Answer 26

For continuity, you can equate the two equations for lim x->1, or
1+a/(1-2) = a+b/sqrt(1)
to get a simplified relation between a and b.

Answer 27

yes i got a= 0 and a+b/1

Answer 28

We do not have a reason to give a=0, we only evaluated each of the left and right-hand limits. We can, however, say that they are equal because of continuity, as I have done above. We have not yet found a and b yet. We need another relation.

Answer 29

hmmm how to get a and b

Answer 30

can you simplify the continuity equation for lim(1-) =lim(1+)?

Answer 31

with two equation

Answer 32

1+a/(1-2) = a+b/sqrt(1)
is the continuity equation for f(1-) = f(1+).
We still have the continuity equation for f'(1-) = f'(1+).

Answer 33

This will give us two equations to solve for a and b.

Answer 34

i dont understand

Answer 35

The question says that the function is continuous AND differentiable.
This implies that the limits both to the left and to the right of x=1 are equal. In simplified form, we say that f(1-) = f(1+).
By evaluating f(x) using the first form given (x=1+a/(x-1)), we get f(1-)=1-a
By evaluating f(x) using the second form given by f(x)=a+b/sqrt(x)), we get f(1+)=a+b
By continuity, f(1-)=f(1+) => 1-a = a+b, => b=1-2a .....(1)
Are you following so far?

Answer 36

b=1-2a
a=-b/2 ...> a=-2a-1/2 ...> 2a+2a=-1..> 4a= -1 ...> a = -1/4

Answer 37

Almost there! You just have to change it to
-a=-b/2 (because of my error in the f'(x), not your fault!) and continue with the rest.

Answer 38

so a=-1

Answer 39

So a=b/2, substitute b=1-2a=1-2(b/2)=1-b => b=1-b => 2b=1 => b=1/2
you can then calculate a from a=b/2=1/4.

Answer 40

b=-3

Answer 41

Want to check starting from:
b=1-2a, and
a=b/2
See if the numbers work?