• Matéria: Matemática
  • Autor: phdjoliveira
  • Perguntado 5 anos atrás

Representa graficamente a função do segundo grau
A) f(x)=7x²-10x
-2

Respostas

respondido por: giovanazilli25
1

Resposta:oi

Explicação passo-a-passo:

respondido por: ThiagoVictorDeMoraes
0

-x/f(x)=0

0=7x²-10x-2

7x²-10x-2=0

ax²+bxc=0

x=-b±√b²-4ac/2a

x=-(-10)±√(-10)²-4×7×(-2)/2×7

-

x=10±√(-10)²-4×7×(-2)/2×7

x=10±√100-4×7×(-2)/2×7

x=10±√100+56/2×7

x=10±√100+56/14

x=10±√156/14

x=10±2√39/14

+-

x=10+2√39/14

x=10-2√39/14

x=5+√39/7

x=10-2√39/14

x=5+√39/7

x=5-√39/7

2

x1=5-√39/7,x2=5+√39/7

x1≈-0,177857,x2≈1,60643

-y,x=0

f(0)=7×0²-10×0-2

0

f(0)=7×0-10×0-2

f(0)=7×0-0-2

f(0)=7×0-2

f(0)=0-2

f(0)-2

xER

f'(x)=d/dx(7x²-10x-2)

d/dx(f+g)=d/dx(f)+d/dx(g)

f'(x)=d/dx(7x²)+d/dx(-10x)-d/dx(2)

f'x=7×2x+d/dx(-10x)-d/dx(2)

f'(x)=7×2x-10-d/dx(2)

f'(x)=7×2x-10-0

f'(x)=14x-10

f(x)=7x²-10x-2,xER

f'(x)=14x-10

f'(x)=14x-10,xER

f'(x)=0

0=14x-10

x=5/7

(-∞,5/7),(5/7,+∞)

x1=0

x2=1

f'(0)=-10

x2=1

f'(0)=-10

f'(1)=4

x<5/7

5/7<x<+∞

x=5/7

f(x)=7x²-10x-2,x=5/7

x=5/7

f(5/7)-39/7

-39/7 em x=5/7

f(x)=7x²-10x-2

f(x)=7x²-10x-2,a=7,b=-10

>0

x

a b x=-b/2a

x=--10/2×7

x

x=5/7

f(x)=7x²-10x-2,x=5/7

f(5/7)=-39/7

-39/7 em x=5/7

lim (f(x)) e lim (f(x))

x→+∞

x→-∞

lim(7x²-10x-2)

x→+∞

lim(7x²-10x-2)

x→-∞

+∞

lim(7x²-10x-2)

x→-∞

+∞

+∞

lim(f(x)/x) e

x→+∞

lim(f(x)/x)

x→-∞

lim x→+∞(7x²-10x-2/x)

lim x→-∞(7x²-10x-2/x)

+∞

x→-∞(7x²-10x-2/x)

+∞

-∞

-∞

x-x

f(-x)=7×(-x)²-10x(-x)-2

f(-x)=7x²-10x(-x)-2

(-)×(-)=(+)

f(-x)=7x²+10x-2

f(-x)=f(x)

f(-x)=-f(x)

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