Sketching $$f(x)=4-x^2$$ and Proving its Concavity in D=[-2,2]

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In summary: Oh no they are not. My mistake. I was looking at another problem I am doing right now. It does not really matter whether they are negative or not for this inequality since the power of 2 makes their difference positive. But yeah they can be negative since they are distinct values of x and the the domain of x is [-2,2].
  • #1
Dostre
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Let D=[-2,2] and $$f:D\rightarrow R$$ be $$f(x)=4-x^2$$ Sketch this function.Using the definition of a concave function prove that it is concave (do not use derivative).

Attempt:
$$f(x)=4-x^2$$ is a down-facing parabola with origin at (0,4). I know that.

Then, how do I prove that f(x) is concave using the definition of a concave function? I got the inequality which should hold for f(x) to be concave:

For two distinct non-negative values of x u and v

$$f(u)=4-u^2$$ and $$f(v)=4-v^2$$

Condition for a concave function:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

After expanding the inequality above I get:

$$(\lambda u-\lambda v)^2\leq(\sqrt{\lambda} u-\sqrt{\lambda} v)^2$$

I do not know what to do next.
 
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  • #2


Dostre said:
Let D=[-2,2] and $$f:D\rightarrow R$$ be $$f(x)=4-x^2$$ Sketch this function.Using the definition of a concave function prove that it is concave (do not use derivative).

Attempt:
$$f(x)=4-x^2$$ is a down-facing parabola with origin at (0,4). I know that.

Then, how do I prove that f(x) is concave using the definition of a concave function? I got the inequality which should hold for f(x) to be concave:

For two distinct non-negative values of x u and v

$$f(u)=4-u^2$$ and $$f(v)=4-v^2$$
Condition for a concave function:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$
You have to show that the above is true, not assume it's true.

Isn't it also required that 0 ≤ λ ≤ 1 ?
After expanding the inequality above I get:

$$(\lambda u-\lambda v)^2\leq(\sqrt{\lambda} u-\sqrt{\lambda} v)^2$$

I do not know what to do next.
 
  • #3


SammyS said:
You have to show that the above is true, not assume it's true.

Isn't it also required that 0 ≤ λ ≤ 1 ?

Yeah it is required that 0 ≤ λ ≤ 1.

But, how do I prove that the below inequality is true?:

$$λ(4−u)^2+(1−λ)(4−v)^2≤4−[(λu+(1−λ)v]^2$$
 
  • #4


Dostre said:
Yeah it is required that 0 ≤ λ ≤ 1.

But, how do I prove that the below inequality is true?:

$$λ(4−u)^2+(1−λ)(4−v)^2≤4−[(λu+(1−λ)v]^2$$
Without Loss of Generality, let u ≥ v .

What else do you know about u & v ?
 
  • #5


SammyS said:
Without Loss of Generality, let u ≥ v .

What else do you know about u & v ?

That they are non-negative. I was working on this problem for a long time and I managed to turn the inequality:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

into

$$\lambda (u-v)^2(1-\lambda)\leq0$$

Since 0<λ<1 the above inequality is true and the function f(x) is concave. I think it is correct.
 
  • #6


Dostre said:
That they are non-negative. I was working on this problem for a long time and I managed to turn the inequality:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

into

$$\lambda (u-v)^2(1-\lambda)\leq0$$

Since 0<λ<1 the above inequality is true and the function f(x) is concave. I think it is correct.
Why do you say u & v are non-negative?
 
  • #7


SammyS said:
Why do you say u & v are non-negative?

Oh no they are not. My mistake. I was looking at another problem I am doing right now. It does not really matter whether they are negative or not for this inequality since the power of 2 makes their difference positive. But yeah they can be negative since they are distinct values of x and the the domain of x is [-2,2].
 

1. What is the graph of $$f(x)=4-x^2$$ in the interval D=[-2,2]?

The graph of $$f(x)=4-x^2$$ in the interval D=[-2,2] is a parabola with its vertex at (0,4) and opening downwards. The graph intersects the x-axis at (-2,0) and (2,0), and the y-axis at (0,4).

2. How do you determine the concavity of a function?

To determine the concavity of a function, you need to find the second derivative of the function. If the second derivative is positive, the function is concave up, and if the second derivative is negative, the function is concave down. If the second derivative is equal to zero, the function does not have a concavity.

3. What is the second derivative of $$f(x)=4-x^2$$?

The second derivative of $$f(x)=4-x^2$$ is -2.

4. How do you prove the concavity of a function?

To prove the concavity of a function, you need to show that the second derivative is always positive or always negative in the given interval. You can do this by taking the second derivative and substituting values from the interval to determine its sign. If it is always positive, the function is concave up, and if it is always negative, the function is concave down.

5. Is $$f(x)=4-x^2$$ a concave function in the interval D=[-2,2]?

Yes, $$f(x)=4-x^2$$ is a concave function in the interval D=[-2,2] because its second derivative is always negative in this interval, indicating that the function is concave down. Therefore, the graph of the function will be in the shape of a downward facing parabola in this interval.

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