Question

mvt

Answer 1

g(x) =
x+2 x smaller than or equal to 1
x^2 x > 1

a)
sketch a graph of the function
i know how to do that

b) show that the function fails to satisfy the hypothesis of the MVT on [-2,2]

c) show that the function does meet the conclusion of the MVT on [-2,2]

Answer 2

justification template for MVT

f is differentiable which implies f is continuous
since f is continuous on [a,b] and differentiable on (a,b) then there is at least one value c, a < c < b such that f'(c) = \((\frac{f(b)-f(a) }{b-a })\)

Answer 3

@tranquility

Answer 4

for b would i just say that it isn't continuous?

Answer 5

f'(c) is derivative at x = c
f(b) - f(a) / b - a is the slope with two points larger and smaller than point c

Answer 6

so also like why is that true

Answer 7

as for part C, mhmm

we just still use

\(f'(c) = \dfrac{f(b)-f(a) }{b-a }\)

and I guess try to find all the values

[-2, 2]
a = -2
b = 2

f(2) = 2^2 = ??
f(-2) = -2 + 2 = ??

and I guess when trying to find f'(c)
we just differentiate both equations and then solve to figure out which one actually works

Answer 8

f'(c) = all that = 1
so f'(c) is equal to 1

Answer 9

yes

Answer 10

differentiate x+2 and x^2 and set it equal to that

Answer 11

derivative of x + 2 is 1
x^2 is 2x
?

Answer 12

oh i'm trying to prove there's a value c where slope is 1

Answer 13

yes

so f'(c) would be 2c or 1 depending on the domain

and so

2c = 1

1 = 1
so on [-2,-1], it's always true

Answer 14

it's because x+2 has a slope of 1 and so any tangent line will have the same slope as the secant line since we found the slope is 1

Answer 15

but why [-2,-1]

Answer 16

because that's the domain of x+2 given in the question

Answer 17

usually these problems have just one point but this has an entire segment

Answer 18

requesting @vocaloid to check if you have some time

Answer 19

hm okay tyty