# Thread: Sum, product and difference of function question

1. ## Sum, product and difference of function question

I am having trouble with the following question.
Two functions are defined as $\displaystyle f(x)=-x^2$ $\displaystyle g(x)=sqrt{x}$

a) Find the smallest possible value of a given that the domain of the function h, where $\displaystyle h(x)=(f+g)(x)$, is a is less than or equal to x and 2 is greater than or equal to x

2. Originally Posted by johnsy123
Find the smallest possible value of a given that....
Means what???

3. I presume that you mean you want to find a so that $\displaystyle a\le x\le 2$ is the domain of $\displaystyle h(x)= -x^2+ \sqrt{x}$.

Okay, what kinds of numbers would NOT give a value for $\displaystyle -x^2+ \sqrt{x}$?

(Hint: you can ignore the $\displaystyle x^2$ term.)

4. Hello, johnsy123!

Very clumsy and misleading wording!

WHO wrote this question?
. . Public flogging may be appropriate punishment.

$\displaystyle \text{Two functions are de{f}ined as: }\:f(x)=\text{-}x^2,\;g(x)=\sqrt{x}$

$\displaystyle \text{Let }h(x) \,=\,(f+g)(x)$

$\displaystyle \text{(a) Given that the domain of }h(x)\text{ is: }\:a \le x \le 2$

. . . .$\displaystyle \text{find the smallest possible value of }a.$

Now that I understand the problem, it's too easy.

We have: .$\displaystyle h(x) \:=\:-x^2 + \sqrt{x}$

We see that $\displaystyle \,x$ must be nonnegative.

So, the domain is: .$\displaystyle 0 \le x \le 2$

. . Therefore: .$\displaystyle a = 0.$ . .Duh!

5. How does a=0? why can't a= 1 or a=2?

All negative numbers.

And i have one more question, if thats okay.
How would i find the range of that function?

6. Originally Posted by johnsy123
How does a=0? why can't a= 1 or a=2?
Because they are not smallest. And your problem asked you to "Find the smallest possible value of a".

Originally Posted by johnsy123
All negative numbers.
??? What is this in response to? There are no negative numbers in the domain of square root if that is what you meant.

Originally Posted by johnsy123
And i have one more question, if thats okay.
How would i find the range of that function?
Your function is $\displaystyle f(x)= -x^2+ \sqrt{x}$ with $\displaystyle 0\le x\le 2$. It is easy to see that the smallest value of f(x) occurs when x= 2: $\displaystyle f(2)= -4+ \sqrt{2}$ but the maximum value is harder. I recommend graphing the function between x= 0 and x= 2.