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VCabral1134 one year ago I need help will give medal and fan!! 5x^3-3 if x<-7 f(x)= 2x^2-3 if -7<_ x<8 evaluate f(x) for x=-7 6+2x if x>_8

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1. anonymous

@AlexanderPhantomhive @Firez @hartnn

2. VCabral1134

@KyanTheDoodle @Michele_Laino @texaschic101

3. VCabral1134

I need help can someone explain this to me?

4. Michele_Laino

there is a jump type discontinuity at x=7

5. VCabral1134

ok what else? I don't understand, I have to find a number @Michele_Laino

6. Michele_Laino

oops.. I meant at x=-7

7. Michele_Laino

we can compute the jump of the function at x = -7, as below: $\begin{gathered} h\left( { - 7} \right) = \mathop {\lim }\limits_{x \to - 7 + } f\left( x \right) - \mathop {\lim }\limits_{x \to - 7 - } f\left( x \right) = \hfill \\ \hfill \\ = 2 \cdot {\left( { - 7} \right)^2} - 3 - 5 \cdot {\left( { - 7} \right)^3} + 3 \hfill \\ \end{gathered}$

8. VCabral1134

ok

9. VCabral1134

do i solve the equation that you provided

10. VCabral1134

@TheSmartOne

11. VCabral1134

is the answer 1,807 ?? @Michele_Laino

12. Michele_Laino

I got 1,813

13. VCabral1134

ok maybe I did something wrong

14. Michele_Laino

we can write this: $\begin{gathered} h\left( x \right) = \left( {2{x^2} - 3} \right) - \left( {5{x^3} - 3} \right) = \hfill \\ \hfill \\ = 2{x^2} - 3 - 5{x^3} + 3 = 2{x^2} - 5{x^3} \hfill \\ \end{gathered}$ now, please try to substitute $$x=-7$$, what do you get?

15. VCabral1134

2* (49) - 3 - 5 * (-343) + 3 98 - 3 - (-1715) +3 95 - (-1712) 1807

16. VCabral1134

@Michele_Laino

17. Michele_Laino

we have: 98-3+1715+3=...?

18. VCabral1134

1813

19. VCabral1134

1813 @Michele_Laino

20. Michele_Laino

correct!

21. VCabral1134

Ok thank you so much!

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