# Thread: Weird inequalities?

1. ## Weird inequalities?

A function is defined by $f(x) = x^2+2x+c$. Find the value of the constant c for which the range of f is given by f(x)≥ 3

2. Originally Posted by Punch
A function is defined by $f(x) = x^2+2x+c$. Find the value of the constant c for which the range of f is given by f(x)≥ 3
You could convert it to turning point form:

$f(x)=(x+1)^2-1+c$

Then you can see that $-1+c$ must equal $3$, so $c$ must equal $4$

3. So this is using the completing the square method

4. Originally Posted by Stroodle
You could convert it to turning point form:

$f(x)=(x+1)^2-1+c$

Then you can see that $-1+c$ must be $\geq 3$, so $c$ must be $\geq 4$
I have a question, the question states that, find the values of the constant c.

now we have an inequality so c can be anything more than 4, how do i answer the question? anything more than 4 or equals to 4 is right?

5. Originally Posted by Punch
I have a question, the question states that, find the values of the constant c.

now we have an inequality so c can be anything more than 4, how do i answer the question? anything more than 4 or equals to 4 is right?
The range is [3, +oo) if c = 4.

6. Originally Posted by mr fantastic
The range is [3, +oo) if c = 4.
Sorry but I didn't get you. If you don't understand my question, this was what i was asking, since c is found to be more than or equals to 4, what should i answer the question "find the value of the constant c"?

7. Sorry Punch, I made an error (I'll fix it now). As Mr Fantastic stated the answer is c=4.

8. Originally Posted by Punch
Sorry but I didn't get you. If you don't understand my question, this was what i was asking, since c is found to be more than or equals to 4, what should i answer the question "find the value of the constant c"?
The answer to the question you posted is c = 4, not $c \geq 4$. That is what my previous reply said. Stroodle made a small misunderstanding of the question: It is required that -1 + c = 3. This misunderstanding should have been clear to you given everything else said in that reply.