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MATH1151 HELP (1 Viewer)

scaryshark09

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use the integral theorem where G = I = F(upper bound) - F(lower bound) and then G' = F'(..) - F'(..) and use chain rule. refer back to ur notes
huh? im so confused

Screenshot 2024-04-01 at 2.11.41 am.png
i used this formula for the first part, but it doesn't work for the others cause of the x^2 and x^3
 

scaryshark09

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Screenshot 2024-04-08 at 3.48.04 pm.png
Does anyone know how to do part (ii) and (iii)??
 

scaryshark09

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Screenshot 2024-04-08 at 3.51.08 pm.pngScreenshot 2024-04-08 at 3.51.29 pm.png

Also, does anyone know how to find a suitable g(x) for these two??
 

scaryshark09

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scaryshark09

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Screenshot 2024-05-05 at 6.13.34 pm.png

does anyone know how to do these two questions? I really would appreciate the help, thanks
 
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liamkk112

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for a), make sure the derivative of the function matches up at x= 9
so n(x-9)^(n-1)sin(1/(x-9)^2) +(x-9)^n cos(1/(x-9)^2) (-2/(x-9)^3) + 2x = 0
then well n =4 is the smallest since otherwise the cos term doesn't go away cos u get something like 1/(x-9), should work

then f'(9) = 0 ? i think

then for c), i think it's also n = 4? not too sure
 

chigurh07

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For part (a) use the definition of a derivative at the point x = 9, so find an expression for f'(9) = lim_{x->9} ((f(x)-f(9)/(x-9)). You'll end up getting the expression lim_{x->9} ((x-9)^(n-1)*sin(1/(x-9)^2)+x+9). This limit only exists if (x-9)^(n-1) approaches 0 since sin(1/(x-9(^2)) will oscillate rapidly as x approaches 9. This means that n > 1 since (x-9)^0 = 1 as x->9, so the smallest integer is 2. For part (b), f'(9) using the limit found in part (a) is 18, since lim_{x->9} ((x-9)^(n-1)*sin(1/(x-9)^2)+x+9) = 0 + 9 + 9 = 18. For part (c) repeat the process used in part (a) and you should get n = 5
 

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