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Calculate a, b, (real numbers) if the function F'= f f = (2x^2 + 1)/sqrt(1+x^2) F=...

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istetz | Student | (Level 2) Honors

Posted September 25, 2010 at 11:23 PM via web

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Calculate a, b, (real numbers) if the function F'= f

f = (2x^2 + 1)/sqrt(1+x^2)

F= (ax + b)*sqrt(1 + x^2)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted September 25, 2010 at 11:27 PM (Answer #1)

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To determine a and b, we'll differentiate F(x).

We notice that F(x) is a product, so we'll differentiate using the product rule.

(u*v) = u'*v + u*v'

u = ax + b => u' = a

v = sqrt(1 + x^2)

Since v is a composed function, we'll use the chain rule to calculate it's derivative.

v' = [1/2sqrt(1+x^2)]*(1+x^2)'

v' = 2x/2sqrt(1+x^2)

We'll simplify:

v' = x/sqrt(1+x^2)

F'(x) = a*sqrt(1 + x^2) + (ax + b)*x/sqrt(1+x^2)

F'(x) = [a(1+x^2) + x(ax+b)]/sqrt(1+x^2)

We know, from enunciation, that:

F'(x) = f(x)

f = (2x^2 + 1)/sqrt(1+x^2)

[a(1+x^2) + x(ax+b)]/sqrt(1+x^2) = (2x^2 + 1)/sqrt(1+x^2)

We'll simplify like terms:

a(1+x^2) + x(ax+b) = 2x^2 + 1

We'll remove the brackets from the left side:

a + ax^2 + ax^2 + bx = 2x^2 + 1

a + 2ax^2 + bx = 2x^2 + 1

The expression from the left side is equal to the expression from the right side, if and only if the coefficients of the correspondent terms are equal.

2a = 2

We'll divide by 2:

a = 1

b*x = 0*x

b = 0

So, F(x) = x*sqrt(1 + x^2), so that F'(x) = f(x).

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william1941 | College Teacher | (Level 3) Valedictorian

Posted September 25, 2010 at 11:46 PM (Answer #2)

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It is given that F'= f.

Now we have F= (ax + b)*sqrt(1 + x^2)

F'= a* sqrt (1 + x^2) + (ax + b)*(2x) / 2 sqrt (1+x^2)

=> a* sqrt (1 + x^2) + (ax + b)*x / sqrt (1+x^2)

=> [a* sqrt (1 + x^2)^2 + (ax + b)*x] / sqrt (1+x^2)

=> [ a (1+ x^2) + (ax + b)*x] / sqrt (1+x^2)

=> [ a + ax^2 +ax^2 + bx]/ sqrt ( 1+ x^2)

As this is equal to (2x^2 + 1) / sqrt(1+x^2)

=> [ a + ax^2 +ax^2 + bx]/ sqrt ( 1+ x^2) = (2x^2 + 1) / sqrt(1+x^2)

=> a + ax^2 +ax^2 + bx = (2x^2 + 1)

=> 2 a x^2 + a + bx = 1 + 2x^2

Now we can equate the coefficients of the terms on the two sides:

=> 2a = 2, b =0, a = 1

So we get b=0 and a=1.

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neela | High School Teacher | (Level 3) Valedictorian

Posted September 25, 2010 at 11:38 PM (Answer #3)

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To calculate real numbers a and b if

f = (2x^2 + 1)/sqrt(1+x^2)..........(1)

F= (ax + b)*sqrt(1 + x^2)...........(2).

We differentiate F(x) = (ax+b)*sqrt(1+x^2)

F'(x) = (ax+b)' *sqrt(1+x^2) +(ax+b)(sqrt(1+x^2)}'

F'= asqrt(1+x^2)+(ax+b) (1/2)(1+x^2)(1/2-1) * (1+x^2)'

F' = asqrt(1+x^2)+(1/2)(ax+b) (2x)/sqr(1+x^2)

F' = asqrt(1+x^2) + (ax+b)x/sqrt(1+x^2) = {a(1+x^2) +ax^2+bx)/sqrt(1+x^2)

F' = {2ax^2+bx+1}/sqrt(1+x^2) ....(3)

So F' = f given. Therefore  the expression at (3) and (1) are identical

So (2ax^2+bx+1)/sqr(1+x^2) and (2x^2+1)/sqri(1+x^2) are identical,

Therefore, 2x^2+bx+1 = 2x^2+1

Therefore, 2a = 2. Or a = 1.

bx = 0 Or b = 0.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted September 26, 2010 at 3:39 AM (Answer #4)

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f= (2x^2+1)/sqrt(1+x^2)

F= (ax+b)*sqrt(1+x^2)

Given F' = f

Now let us differencite F:

Let F = u*v such that:

u = ax+b    ==>  u' = a

v= sqrt(1+x^2) ==> v' = x/sqrt(1+x^2)

==> F' = u'v + uv'

            = a*sqrt(1+x^2) + (ax+b)*2x /sqrt(1+x^2)

             = [a(1+x^2) + x(ax+b) ]/sqrt(1+x^2)

              = [ a+ ax^2 + ax^2 + bx] /sqrt(1+x^2)

                = (2a)x^2 + (b)x + a]/sqrt(1+x^2)

But F'= f

==> 2ax^2 + bx + a = 2x^2 + 1

==> a= 1

==> b= 0

==> F(x) = x*sqrt(1+x^2)

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