Calc- Test Review

Use the IVT to show that the functions f(x)= has a zero on the interval [-2,-1]

1) f(x) is cont. because poly functions are cont.

2) f(-2) = 2 f(-1) = -4

Since f(x) is cont. and f(-2) ⟨ f(c) ⟨ f(-1), then ∃ some C∃ (-1,-2) s.t f(c) = 0
Discontinuity at x= (first)

x= 8

Removable (bc x-8 cancels)
Discontinuity at x= (second)
- x= -1
- Non-Removable (stays in denominator)
Sketch the graph of any function

lim f(x) = 4 and lim f(x) = -1
x→ 2+ x→ 2-

Is the function cont.?
- No, the limit DNE
- (x--> c = x-axis
- f(x)=b = y axis)
sinx/x = y1
sin3x/x = y2

(Must be going to 0)
- y1 = 1
- y2 = 3
1-cosx = y

(Must be going to 0)

y = 0
7 + 2x - 5x²
--------------
2x² + 3x - 4

x--> - ∞

-5/2

When top and bottom are both x² heavy you take the coefficient and divide.
=

= 0 -3

= -3

(Anything to the -∞ is 0 except 1 and 0)
=

= -∞ +2

= -∞

(e^∞ is ∞, e^-∞ is -∞)

Author

GoBroncos

ID

356577

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Calc- Test Review

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Updated

2021-10-19T19:18:06Z