# Max point of a function in $x$ range $0$ to $4$ and inflection points

here's the function: $f(t) = -t^3 + 7t^2 + 200t$

how can I calculate the max $y$ value of a function for $x\in [0,4]$ range? Trying one-by-one manually?

And, I've tried to calculate the gobal min and max: $-6.158, 10.82$, and then the inflection point: $7/3$, so, this inflection point should be the max point between $-6.158$ and $10.82$, right? but it's not, because when I test the function with $x=3$, I get a greater value that what I get when setting $x = 7/3$. Why?

Thank you!

For this problem, you would look at the function values at the endpoints and at the critical points you found to see what is happening because the critical points you correctly found are outside the range you were given to investigate.

We have:

$f(t) = -t^3 + 7t^2 + 200t$

$f(0) = 0$

$\displaystyle f\left(\frac{7}{3}\right) = \frac{13286}{27} = 492.0740$

$f(4) = 848$.

So we have a min at $0$ and a max at $4$.

If we look at a plot, lets verify this - see the min at $0$ and the function is increasing on the range $[0,4]$. You could have also explored what happens to the derivative in the given range.

• Thanks, so the best way to check the max in this range is by investigation. But what about the inflection point 7/3? I can see that this function has a constant slop with x from 0 to 4.. Thank you again. – user71331 Apr 6 '13 at 23:13
• What does the first derivative tell you over the range $[0, 4]$? Is it positive the entire time? The derivative you found was giving you some global information, but it is not very helpful when you limit to a specific range. – Amzoti Apr 6 '13 at 23:15
• So my inflection point is in a global range, not so precise? – user71331 Apr 6 '13 at 23:20
• I wouldn't say not so precise, just not that useful. Note that above, we did use it to see what happens there, so it did tell us something, that it was not the min or max as the function is still increasing there over this limited range. Don't over think it. – Amzoti Apr 6 '13 at 23:25
• Ok, thank you very much Amzoti, you helped me a lot today :D – user71331 Apr 6 '13 at 23:31