f(x)=−2x3−5x2−5x−6x+2 when x<−2x<−2, f(x)=−5x2+6x+a when −2≤x−2≤x.

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AngeNova

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f(x)=−2x3−5x2−5x−6x+2 when x<−2x<−2, f(x)=−5x2+6x+a when −2≤x−2≤x.

I'm doing my web homework, and just can't figure out how to solve this problem...

The function [FONT=MathJax_Math-italic]ff[/FONT] is given by the formula
[FONT=MathJax_Math-italic]f​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT]
​
[/FONT]​
when [FONT=MathJax_Math-italic]x[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT]x<−2[/FONT] and by the formula
[FONT=MathJax_Math-italic]f​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]a[/FONT]
​
[/FONT]​
when [FONT=MathJax_Main]−[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math-italic]x[/FONT]−2≤x[/FONT].
What value must be chosen for [FONT=MathJax_Math-italic]aa[/FONT] in order to make this function continuous at -2?

I have tried to equal them.. but I guess I did something wrong.

Next problem.

 If a rock is thrown into the air on small planet with a velocity of [FONT=MathJax_Main]23[/FONT] meters/second, its height in meters after t seconds is given by [FONT=MathJax_Math-italic]y[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]23[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4.9[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]2[/FONT]. Find the velocity of the rock when [/FONT]t=1:

I have put t=1 in the formula, but..it isn't good :(
THE RESULT IS NOT 18.1

Next problem

A particle moves along a straight line and its position at time [FONT=MathJax_Math-italic]tt[/FONT] is given by [FONT=MathJax_Math-italic]s[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]24[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]42[/FONT][FONT=MathJax_Math-italic]t[/FONT] where s is measured in meters and t in seconds.
[/FONT]Find the velocity of the particle at time [FONT=MathJax_Math-italic]t[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][/FONT]: 42 meters/second.
The particle stops moving (i.e. is in a rest) twice, once when [FONT=MathJax_Math-italic]t[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]A[/FONT]t=A[/FONT] and again when [FONT=MathJax_Math-italic]t[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT] where [FONT=MathJax_Math-italic]A[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT].
[FONT=MathJax_Math-italic]A[/FONT] is 1 seconds,
and [FONT=MathJax_Math-italic]B[/FONT] is 7 seconds.
What is the position of the particle at time [FONT=MathJax_Main]16[/FONT]? 2720meters.
Finally, what is the TOTAL distance the particle travels between time [FONT=MathJax_Main]0[/FONT] and time [FONT=MathJax_Main]16[/FONT]?

EVERYTHING IS GOOD EXCEPT FOR THE LAST question. I CAN'T FIGURE OUT HOW TO CALCULATE THE TOTAL DISTANCE. P.S. IT'S NOT 2720 NEITHER


Thank you a lot for helping! :D
 
Last edited:
AngeNova said:
I'm doing my web homework, and just can't figure out how to solve this problem...

The function [FONT=MathJax_Math-italic]ff[/FONT] is given by the formula
[FONT=MathJax_Math-italic]f​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT]
​
[/FONT]​
when [FONT=MathJax_Math-italic]x[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT]x<−2[/FONT] and by the formula
[FONT=MathJax_Math-italic]f​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]a[/FONT]
​
[/FONT]​
when [FONT=MathJax_Main]−[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math-italic]x[/FONT]−2≤x[/FONT].
What value must be chosen for [FONT=MathJax_Math-italic]aa[/FONT] in order to make this function continuous at -2?

I have tried to equal them.. but I guess I did something wrong. You need to equate the expressions after substituting x = -2. Try again!
Next problem.

 If a rock is thrown into the air on small planet with a velocity of [FONT=MathJax_Main]23[/FONT] meters/second, its height in meters after t seconds is given by [FONT=MathJax_Math-italic]y[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]23[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4.9[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]2[/FONT]. Find the velocity of the rock when [/FONT]t=1:

I have put t=1 in the formula, but..it isn't good :(
THE RESULT IS NOT 18.1 If you put t=1 into the formula for height, you get height, so the height is 18.1 m.
You want velocity. So what is the velocity function? (Hint: dy/dt) Have another go.

Next problem

A particle moves along a straight line and its position at time [FONT=MathJax_Math-italic]tt[/FONT] is given by [FONT=MathJax_Math-italic]s[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]24[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]42[/FONT][FONT=MathJax_Math-italic]t[/FONT] where s is measured in meters and t in seconds.
[/FONT]Find the velocity of the particle at time [FONT=MathJax_Math-italic]t[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][/FONT]: 42 meters/second.
The particle stops moving (i.e. is in a rest) twice, once when [FONT=MathJax_Math-italic]t[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]A[/FONT]t=A[/FONT] and again when [FONT=MathJax_Math-italic]t[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT] where [FONT=MathJax_Math-italic]A[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT].
[FONT=MathJax_Math-italic]A[/FONT] is 1 seconds,
and [FONT=MathJax_Math-italic]B[/FONT] is 7 seconds.
What is the position of the particle at time [FONT=MathJax_Main]16[/FONT]? 2720meters.
Finally, what is the TOTAL distance the particle travels between time [FONT=MathJax_Main]0[/FONT] and time [FONT=MathJax_Main]16[/FONT]?

EVERYTHING IS GOOD EXCEPT FOR THE LAST question. I CAN'T FIGURE OUT HOW TO CALCULATE THE TOTAL DISTANCE. P.S. IT'S NOT 2720 NEITHER

Make sure you know the difference between displacement and distance. At t=1 and t=7, the particle is actually changing direction. You need to calculate the distance walked from t=0 to t=1. Then the distance from t=1 to t=7 (remember distance is positive). Then the distance from t=7 to t=16. Then add all these positive values together to get total distance. Give it a go.

Thank you a lot for helping! :D
Click to expand...
See comments in red.
 
Thank you! got the second problem right!

I didn't write the first problem well, sorry :(
I'll post it again.


[FONT=&quot]The function [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Math-italic]f[/FONT][/FONT][FONT=&quot] is given by the formula[/FONT][FONT=&quot][/FONT]
[FONT=&quot]
[FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT]
​
[/FONT]​
[FONT=&quot]when [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][/FONT][FONT=&quot] and by the formula [/FONT]
[FONT=&quot][/FONT]
[FONT=&quot]
[FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]a[/FONT]
​
[/FONT]​
[FONT=&quot]when [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math-italic]x[/FONT][/FONT][FONT=&quot] [/FONT]
[FONT=&quot]What value must be chosen for [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Math-italic]a[/FONT][/FONT][FONT=&quot] in order to make this function continuous at -2?


I'm working on the last one! :)[/FONT]
 
Reply

I did the second one! Still doing the third one.

I didn't give the complete exercise for the first one. Here it is again.

[FONT=&quot]The function [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Math-italic]f[/FONT][/FONT][FONT=&quot] is given by the formula[/FONT][FONT=&quot][/FONT]
[FONT=&quot]
[FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT]
​
[/FONT]​
[FONT=&quot]when [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][/FONT][FONT=&quot] and by the formula [/FONT]
[FONT=&quot][/FONT]
[FONT=&quot]
[FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]a[/FONT]
​
[/FONT]​
[FONT=&quot]when [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math-italic]x[/FONT][/FONT][FONT=&quot]. [/FONT]
[FONT=&quot]What value must be chosen for [/FONT][FONT=&quot][/FONT][FONT=&quot][FONT=MathJax_Math-italic]a[/FONT][/FONT][FONT=&quot] in order to make this function continuous at -2?[/FONT]
 
AngeNova said:
The function [FONT=MathJax_Math-italic]f[/FONT] is given by the formula
[FONT=MathJax_Math-italic]f​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT]
​
[/FONT]​
when [FONT=MathJax_Math-italic]x[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][/FONT] and by the formula
[FONT=MathJax_Math-italic]f​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]a[/FONT]
​
[/FONT]​
when [FONT=MathJax_Main]−[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math-italic]x[/FONT][/FONT].
What value must be chosen for [FONT=MathJax_Math-italic]a[/FONT] in order to make this function continuous at -2?
Click to expand...
What is the value of the first formula when x = -2? What is the value of the second formula when x = - 2? What must "a" be, in order to have the second formula's value equal the first formula's?

Please be complete. Thank you! ;)
 
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