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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).

$$f(x)=4-x^2$$ 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.

**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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