Intermediate Value Theorem
Define the Intermediate Value Theorem
If f(x), a function, is continuous on the closed interval [a,b], and x is a number between f(a) and f(b), then there must exist one c such that f(c)=x.
Define the Mean Value Theorem
If f(x) is continuous over the closed interval [a,b] and differentiable on the open interval (a,b), then there must exist a value c on (a,b) such that f’(c)= f(b)-f(a)/(b-a).
Define the Extreme Value Theorem
If f(x) is continuous over the closed interval [a,b] then there must be both an absolute maximum and absolute minimum value on the interval.
Define the Squeeze Theorem
If g(x)<f(x)<h(x), and lim x→c of g(x) = lim x→c of h(x), then, lim x→c of f(x) = lim x→c of g(x) = lim x→c of h(x).
Define Rolle's Theorem
If f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) then f(a)=f(b), then there exists one point, c, on the open interval (a,b) where the derivative f’(c)=0.
On the interval (5,10), if f(5)= -10 and f(10) = 20, is there a value, c, where f(c)= 6?
Yes. Due to the IVT, there is a value, c, where f(c)= 6, because f(5) = -10 and f(10)= 20
On the interval (3,9), if g(3)= 6 and g(9)= 30, is there a value, c, where g’(c)= 4?
Yes. Due to the MVT, because the AROC of the interval (3,9) is equal to 4, there is a value, c, where g’(c) = 4.
(30-6)÷(9-3) = 24÷6 = 4
Find the extreme on 𝑓(𝑥) =8x^3 +81x^2 −42x −8 on [−8,2].
Absolute Maximum: 1511 at x=−7
Absolute Minimum: −13.3125 at x= .25
(find the CV, consider your end points, find the max and min)
-x^2 is less than g(x) is less than x^2. What is the limit of g(x) as x approaches 0?
0, as -x^2 goes to 0 and x^2 goes to 0, g(x) is "squeezed" to 0.
See (Red1) in Desmos
x=-2
f(x)=3x^2+4x+9 On the interval (3,7), is there a number, c, where f(c)= 47?
No. Due to the IVT, there is no value, c, on the interval (3,7) where f(c)=47, because f(3)= 48 and f(7)= 184.
g(x)=4x^3 - 2x^2 + 5x - 7
On the interval (-2,4), is there a value, c, where g’(c) = 49?
Yes. Due to the MVT, because the AROC of the interval (-2,4) is equal to 49, there is a value, c, where g’(c)=49.
g(-2)=-57
g(4)= 237
(237-(-57))÷(4-(-2))=294 ÷ 6 = 49
Find global min and max for f(𝑥) =(2−8x)^4(x^2−9)^3 on [−3,3]
Absolute max: 0 when x= -3, .25, 3
Absolute min: -1.36x10^7 x= -1.8239
x→0 Lim (∣x∣⋅sin(1/x^2))
0
-1<sin(1/x^2)<1 −∣x∣≤∣x∣sin(1/x^2)≤∣x∣
lim x→∣x∣ =0 lim x→-∣x∣ =0
See (Orange2) in Desmos
x=3
If f(x)=3-x^2; [-2,1], where k=0, determine if the IVT holds true for the value k.
Yes. f(x) is continuous over the closed interval, f(-2) does not equal f(1), k=0 between f(-2) and f(1). Therefore, the IVT applies and such value c exists that f(c)=0.
Suppose (f(x) is continuous and differentiable on [-7,0], f(-7) = -3, and f'(x)≤ 2 for all x in the interval. What is the largest possible value for f(0)?
f(0) = 11 Because of the MVT, you can set the AROC on the interval equal to 2, then solve algebraically for f(0).
(f(0)-f(-7))÷(0-(-7))=
(f(0) + 3))÷(0+7) ≤ 2
f(0) + 3 = 2*7
f(0)=14-3
f(x)=ln(x^2+4x+14); [-4,2]
Max: 3.2581 when x=2
Min: 2.3026 when x=-2
5 ≤ 𝑓(𝑥) ≤ x^2 + 6x − 2
As lim x→1 f(x)
Lim x→1 f(x)=5
See (Green3) in Desmos
x=-0.88929
x=0.88929
f(x)=1/x; [2,5] where k=-1. determine if the IVT holds true for the value k.
No. F is continuous over the closed interval, f(2)= ½ and f(5)= ⅕ however the value k=-1 is not between f(2) and f(5).
In this piecewise function, f(x)= {x^2 +1; -1≤x≤0} {-x+1; 0≤x≤1}, is the MVT applicable?
No. f(x) is continuous, because the limit as x approaches 0 from the left and right equals 1. However, f(x) is not differentiable as the derivative from the left and the right are not equal to each other, thus the MVT is not applicable.
f(x)= (x+4)/(2x^2+x+8); [-10,0]
Max: .5 when x=0
Min: -.03128 when x=-4-3sqrt2
Lim x→0 x^2sin (1/x)
Lim x→0 x^2sin (1/x)=0
Find f’(0) on csc(pi/4)*4x^2-6
(No Graph)
x=0 y=-6