Hi! This is piecewise equations! Please help if you can :)

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Notice that f(x) is only defined over the intervals 0 ≤ x < 3 and 3 < x ≤ 6; that is, all real numbers x between 0 and 3, and between 3 and 6, but not including 3. Since there is no definition for f(x) at x = 3, the value of f (3) is undefined.

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givi [52]

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Click here please. The problems are in picture format

padilas [110]

First picture:

You got the first two right (good job!), so I'll just tackle the third one. If you have to compute $f(x-4)$ , it means that you have to substitute $x-4$ in place of $x$ . Differently from the first two points, this will generate a new function, rather than a specific value:

$f(x) = \sqrt{x+4}+2 \implies f(x-4) = \sqrt{(x-4)+4}+2 = \sqrt{x}+2$

Second picture:

Given the point $(x,y)$ highlighted in the picture, you can deduce that the base is $2x$ units long (since it spans from $(-x,0)$ to $(x,0)$ ) and the height is $y$ units long (because it spans from $(x,0)$ to $(x,y)$ ). So, the area of the rectangle is the multiplication between base and height:

$A = 2xy$

But we know that $y=f(x)=\sqrt{36-x^2}$ , so we have

$A = 2x\sqrt{36-x^2}$

The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have

$36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6$

But since the problem is symmetric, the answer is

$0 \leq x \leq 6$

You can only see the answer $0 < x < 6$ because, if you choose $x=0$ or $x=6$ , the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently