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Oct 14, 2021, 7:17:03 AM10/14/21

to tmoz

Ilgilenen arkadaşlara teşekkürler.

20211014_141345.jpg

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Oct 14, 2021, 11:05:31 AM10/14/21

to TMOZ

f(x) =(x^2-x-6)/(x-3) olduğundan

x=/3 için f(x)=x+2'dir.

limit x->3 f(x)=5

f'(x)=x-2

f'(x)=(x-2)*(x-5)/(x-5)=(x^2-7x+10)/(x-5)

a=7

b=10

c=-5

a*b-c=70+5=75

On Thursday, October 14, 2021 at 2:17:03 PM UTC+3 madmai wrote:

Ilgilenen arkadaşlara teşekkürler.

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