Respuesta :
Answer:
27
Step-by-step explanation:
In this case, because we have a polynomial where all x-values exist, we can plug the value (x=2) straight into the function:
[tex]\lim_{x \to \ 2} (2x^3+x^2+7)[/tex]
[tex]= 2(2)^3+(2)^2+7[/tex]
[tex]= 2(8)+4+7[/tex]
[tex]= 27[/tex]
The limit of (2x^3+x^2+7) as x approaches 2 equals 27, which makes sense since the value of (2x^3+x^2+7) at x = 2 is 27.
Answer:
Looking Ahead to Calculus: Lessons 1-5 Quick Check Answers
Lesson 1
Q1 B. -21
Q2 D. 4
Q3 D. 4 (yes, again)
Q4 B. lim f(x) = 5 [x→3+] ;
D. lim f(x) = 6 [x→3-] ;
F. lim f(x) [x→3] does not exist...
Lesson 2
Q1 A. lim x = a [x→a] ;
B. lim a = a [x→a];
D. lim x = 5 [x→5]
Q2 D. 16
Q3 B. 10a
Q4 D. 3
Lesson 3
Q1 A. -1
Q2 D. 1
Q3 B. It is discontinuous because there is a value a for which f(a) is not defined.
Q4 D. It is discontinuous because there is a value a such that lim f(x) [x→a] does not exist.
Lesson 4
Q1 D. 3
Q2 A. -13
Q3 C. y = -2x + 6
Q4 C. f'(a) = 8a + 2
Lesson 5
Q1 C. 2
Q2 D. v(t) = 8t - 2
Q3 B. 902 mi./min.
Q4 A. The car is slowing down at a rate of 10 mi./h^2.
Have a great summer
