c3 maths functions

given that g(x) = 2x^2-3 and h(x)=√5x+2, find expression for gh(x). x ≥ -2/5 solve inequality gh(x) ≥ x.

for gh(x) i did 2(√5x+2)^2-3 to get 10x+1, not sure if that is correct? and im unsure of how to do the inequality part. Can anyone explain?

Reply 1

ghostwalker

17

Is that:

h(x)=\sqrt{5}x+2

or

h(x)=\sqrt{5x}+2

or

h(x)=\sqrt{5x+2}

?

If you mean the last one then your gh(x) is correct, otherwise not.

Reply 2

AndyOC

OP

2

Original post by ghostwalker

Is that:

h(x)=\sqrt{5}x+2

or

h(x)=\sqrt{5x}+2

or

h(x)=\sqrt{5x+2}

?

If you mean the last one then your gh(x) is correct, otherwise not.

yeah its the last one

Reply 3

AndyOC

OP

2

Original post by ghostwalker

Is that:

h(x)=\sqrt{5}x+2

or

h(x)=\sqrt{5x}+2

or

h(x)=\sqrt{5x+2}

?

If you mean the last one then your gh(x) is correct, otherwise not.

for the inequality do i just substitute x=-2/5 in gh(x)?

Reply 4

Dalek1099

21

Original post by AndyOC

for the inequality do i just substitute x=-2/5 in gh(x)?

treat the inequality like an equation but be careful not to divide by -1(if you do flip the signs).

Reply 5

ghostwalker

17

Original post by AndyOC

for the inequality do i just substitute x=-2/5 in gh(x)?

No. x>= -2/5 is required for your square root ( h(x) ) to be valid, and is just a restriction on x.

You need to solve gh(x) >= x

Substitute what you've worked out for gh(x) and then solve the inequality.

Reply 6

AndyOC

OP

2

Original post by ghostwalker

No. x>= -2/5 is required for your square root ( h(x) ) to be valid, and is just a restriction on x.

You need to solve gh(x) >= x

Substitute what you've worked out for gh(x) and then solve the inequality.

ah yeah i see, understand it now. Thanks for the help :smile:

c3 maths functions

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