Why is my answer incorrect?

[gusty venture]

This is what i did:

[agile schooner]

Is it necessary for $\min(f(x))$ to be the value where $x$ is the least? And vice versa for $\max(f(x))$ 😉

[clever cove]

Marianne

[agile schooner]

Looking at your graph, for example, $f(-5)=4,$ which is actually greater than all of the other values in the range of $f(x).$

[clever cove]

Marianne

[gusty venture]

@agile schooner that was the upper bound for f(x)

h(-5) will be the lower bound since f(-2) is the lower bound for f(x)

h(-8) will be the upper bound because f(-5) is the upper bound for f(x)

idk what you mean by "is it necessary for min(f(x)) to be the value for x is the least"

because i did not say that f(-5) will give the lower bound i said f(-2) will

[agile schooner]

@gusty venture Ah, I see. Sorry, I interpreted it the wrong way. I also think your answer is correct. If your teacher is grading your work, I think you should ask them to explain.

[gusty venture]

my teacher is not grading the work i have to input it into a grading software. maybe i am entering it wrong or maybe my answer is just incorrect. but thx anyways

[agile schooner]

I see. Then maybe you should ping helpers and see if they can spot your error.

[full dune]

The domain you wrote is that of f(x), not h(x).

[gusty venture]

@full dune wait why is the domain not the same for both functions?

[full dune]

f(x) is defined for x in [-5,3].
If f(x+3) had the same domain, then f(3+3) = f(6) would be defined, but 6 isn't in the domain of f...

By double checking your work, notice how x=-8 would minimize h(x), but you just wrote that x has to be in [-5,3], so something must be afoot

[gusty venture]

oh so the domain would just be -8<=x<=0

i see thx i intuitively just thought the domain would be the same

thx so much

.close