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CÁLCULO DIFERENCIAL E INTEGRAL I - Conevyt

CÁLCULO DIFERENCIAL E INTEGRAL I - Conevyt

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2. Si<br />

f ( x)<br />

f ( x + h)<br />

− f ( x)<br />

lim<br />

h 0 h<br />

= →<br />

3<br />

f ( x)<br />

= 5x<br />

− 3x<br />

+ 2<br />

(2)<br />

f (x + h) = 5 (x + h) 3 – 3 (x + h) + 2 (3)<br />

Sustituyendo 2 y 3 en 1 tenemos.<br />

f(<br />

x)<br />

= lim<br />

h→0<br />

5(<br />

x<br />

+ h)<br />

3<br />

−<br />

3(<br />

x<br />

+ h)<br />

+ 2 − ( 5x<br />

h<br />

3<br />

− 3x<br />

+<br />

Efectuando las operaciones indicadas nos queda.<br />

f(<br />

x)<br />

f(<br />

x)<br />

f(<br />

x)<br />

f(<br />

x)<br />

f(<br />

x)<br />

= lim<br />

h→0<br />

= lim<br />

h→0<br />

= lim<br />

h→0<br />

=<br />

h→0<br />

5x<br />

3<br />

15x<br />

2<br />

2<br />

3<br />

+ 15x<br />

h + 15xh<br />

+ 5h<br />

− 3x<br />

− 3h<br />

+ 2 − 5x<br />

h<br />

+ 3x<br />

− 2<br />

2<br />

h ( 15x<br />

lim 15x<br />

=<br />

h→0<br />

lim 15x<br />

2<br />

2<br />

h + 15xh<br />

h<br />

2<br />

2<br />

− 3h<br />

+ 15xh<br />

− 3)<br />

h<br />

+ 15x(<br />

0)<br />

+<br />

− 3<br />

5(<br />

0)<br />

3<br />

-La razón de cambio de 5x es<br />

-La razón de cambio de -3x es -3<br />

2<br />

− 3<br />

′ x<br />

76<br />

2)<br />

2<br />

de donde f ( x)<br />

= 15 − 3<br />

2<br />

15x<br />

3<br />

(1) entonces

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