Question

Find the largest value of A such that the function f(t)=(t^3)+(2t^2)-(5t)+7 is decreasing for all t in the interval (0,A).

Answer 1

I have no idea how to do this

Answer 2

do you know how to find where a function is increasing or decreasing?

Answer 3

ive honestly been kind of lost the past few days in my math class.

Answer 4

a nice explanation would be well received :)

Answer 5

All right . now a function is said to be increasing if in a range it shows positive slope and decreasing if negative.
slope at any point is just the derivative of the function here df(t)/dt
The funtion changes from decreasing to increasing once the slope at any point becomes zero (unless t is point of inflection where it donot change from -ve to positive or +ve to negative)
so inorder to find the range (0,A) we need to find the value of A where the f'(A) = 0
this means 3t^2+4t+-5=0(substitute A for t)
this leave us only one positive value of A = 1/3 * ((19)^1*2 -2)

Answer 6

this leave us only one positive value of A = 1/3 * ((19)^1*2 -2)

explain how you got those numbers please

Answer 7

sorry it is sqrt(19) just solve the quadratic

Answer 8

\[f(t)=t^3+2t^2-5t+7\] is decreasing where
\[f'(t)=3t^2+4t-5<0\] so this is what we have to solve

Answer 9

quadratic formula gives the zeros at
\[\frac{-2\pm\sqrt{19}}{3}\] and since this is a parabola that faces up, it will be negative between the zeros and positive outside them, so it is negative over the entire interval
\[(\frac{-2\sqrt{19}}{3},\frac{-2+\sqrt{19}}{3})\]

Answer 10

therefore the answer to your question is

\[A=\frac{-2+\sqrt{19}}{3}\]

Answer 11

im not getting that answer when i use the quadratic equation

Answer 12

oh ... plase try again and verify steps I got the same agian. I hope you understood the way .Answer is not so important if you know how to

Answer 13

i know how to use the quadratic formula but im not getting the same numbers as you all.

Answer 14

show work?

Answer 15

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-4\pm\sqrt{4^2-4\times 3\times (-5)}}{2\times 3}\]
\[x=\frac{-4\pm\sqrt{76}}{6}\]
\[x=\frac{-4\pm\sqrt{19}}{6}\]
\[x=\frac{-2\pm\sqrt{19}}{3}\]

Answer 16

typo there, should have been
\[x=\frac{-4\pm2\sqrt{19}}{6}\]

Answer 17

part you might have missed was
\[\sqrt{76}=\sqrt{4\times 19}=2\sqrt{19}\]

Answer 18

why does the 6 in the denominator become a 3????