Answer :
Answer:
[tex] f(x+h) = 3(x+h)^2 +(x+h) +2= 3(x^2 +2xh+h^2) +x+h+2[/tex]
[tex]f(x+h) = 3x^2 +6xh +3h^2 +x+h+2= 3x^2 +6xh +x+h+ 3h^2 +2[/tex]
And replacing we got:
[tex] lim_{h \to 0} \frac{3x^2 +6xh +x+h+ 3h^2 +2 -3x^2 -x-2}{h}[/tex]
And if we simplfy we got:
[tex] lim_{h \to 0} \frac{6xh +h+ 3h^2 }{h} =lim_{h \to 0} 6x + 1 +3h [/tex]
And replacing we got:
[tex]lim_{h \to 0} 6x + 1 +3h = 6x+1[/tex]
And the bet option would be:
C. 6x + 1
Step-by-step explanation:
We have the following function given:
[tex] f(x) = 3x^2 +x+2[/tex]
And we want to find this limit:
[tex] lim_{h \to 0} \frac{f(x+h) -f(x)}{h}[/tex]
We can begin finding:
[tex] f(x+h) = 3(x+h)^2 +(x+h) +2= 3(x^2 +2xh+h^2) +x+h+2[/tex]
[tex]f(x+h) = 3x^2 +6xh +3h^2 +x+h+2= 3x^2 +6xh +x+h+ 3h^2 +2[/tex]
And replacing we got:
[tex] lim_{h \to 0} \frac{3x^2 +6xh +x+h+ 3h^2 +2 -3x^2 -x-2}{h}[/tex]
And if we simplfy we got:
[tex] lim_{h \to 0} \frac{6xh +h+ 3h^2 }{h} =lim_{h \to 0} 6x + 1 +3h [/tex]
And replacing we got:
[tex]lim_{h \to 0} 6x + 1 +3h = 6x+1[/tex]
And the bet option would be:
C. 6x + 1