P1 Maths Question, Help.

I'd be gratefull if anyone could show me how to do this question:

f(x) = x^2 - kx + 9, where k is a constant

a) Find the set of values of k for which the equation f(x) = 0 has no real solutions.

Given that k = 4

b) Express f(x) in the form (x-p)^2 + q, where p and q are constants to be found.

c) Write down the minimum value of f(x) and the value of x for which this occurs.

I've really forgotten how to do it.

Reply 1

lgs98jonee

2

Bebop

I'd be gratefull if anyone could show me how to do this question:

f(x) = x^2 - kx + 9, where k is a constant

a) Find the set of values of k for which the equation f(x) = 0 has no real solutions.

Given that k = 4

b) Express f(x) in the form (x-p)^2 + q, where p and q are constants to be found.

c) Write down the minimum value of f(x) and the value of x for which this occurs.

I've really forgotten how to do it.

a.
(b^2-4ac) must be positive to give real solution so,
therefore b^2>=36
so b>=¦6¦ (modulus 6 i think) [to give realy] so u want -6<k<6 to give no real solution :-)
b.complete square? (x-2)^2-4+9=(x-2)^2+5
c. differentitate
d(f(x))/dx=2x-4
when 2x-4=0, x=2

Reply 2

Euler

3

Bebop

I'd be gratefull if anyone could show me how to do this question:

f(x) = x^2 - kx + 9, where k is a constant

a) Find the set of values of k for which the equation f(x) = 0 has no real solutions.

Given that k = 4

b) Express f(x) in the form (x-p)^2 + q, where p and q are constants to be found.

c) Write down the minimum value of f(x) and the value of x for which this occurs.

I've really forgotten how to do it.

Its a quadratic equation with
a = 1
b = -k
c = 9

a quadratic equation has no real root if

b^2 < 4ac

therefore, substituting gives:

(-k)^2 < 4*1*9

K^2 < 36

K < plus OR minus 6

so inequality is :

-6 < k <6

For second part use completing the square method to get:

(x - 2)^2 + 5

For the third part differentiate the equation to get:

2x - 4

equate this to zero to get:

x = 2

which gives f(x) = 5

differentiate again to get 2, which is greater than 0, therefore their is only one minimum value for the function.

Reply 3

Datura

OP

1

Thank you both very, very much!

Well since I'm brushing up on the p1 I might as well get another thing cleared up.

In integration I've never felt confident knowing what to do when you have to integrate a term containing a real number divided by the root of x.

Eg. 3√x + 12/√x

I know that the root of x is equvalent to x to the power a half.

Anyway I'd appreciate any ideas on this.

Reply 4

john !!

14

f'(x) = 3√x + 12/√x
= 3x^1/2 + 12x^-1/2

f(x) = 2x^3/2 + 24x^1/2 + c

I think.

Reply 5

elpaw

15

Bebop

Thank you both very, very much!

Well since I'm brushing up on the p1 I might as well get another thing cleared up.

In integration I've never felt confident knowing what to do when you have to integrate a term containing a real number divided by the root of x.

Eg. 3√x + 12/√x

I know that the root of x is equvalent to x to the power a half.

Anyway I'd appreciate any ideas on this.

like you said, √x is the x to the half, so just treat it as x^n with n=1/2, as you would any other integral of powers of x.

ie ∫√x dx = ∫x^1/2 dx = (x^3/2)/3/2 +C = 2/3 x√x + C
and ∫(dx/√x) = ∫x^-1/2 dx = (x^1/2)/1/2 +C = 2√x +C

Reply 6

Datura

OP

1

Thank you, I'm extremely grateful to you all.

P1 Maths Question, Help.

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