Question

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Answer 1

Answer:

1)Function is continuous in [1, 4] as it is a polynomial function and polynomial function is always continuous.

2)   f'(x)=2x-4 exists in [1, 4], hence derivable. Conditions of MVT theorem are satisfied, hence there exists, at least one  such that c belongs to (1,4)

Step-by-step explanation:

f'(c)=f(4)-f(1)/4-1

f(4)= -3

f(1)= -6

f'(c)=2c-4

2c-4=1

c=5/2

Answer 2

• mean value theorem

if a function f(X) is

• continuous in [ a,b]
• derivable in (a,b)

then there exists at least one real number c in (a,b) such that f'(c)

$= \frac{f(b) - f(a)}{b - a}$

step by step explanation

f( X) = x²-4x-3

in [ 1,4]

• since a polynomial function is everywhere continuous and differentiable .
• therefore ,f(X) I continuous on [ 1,4] and differentiable in (1,4) .

⏩by mean value theorem

there must exist one real no belongs to (1,4).

such that , f'(C)

$= \frac{f(4) - f(1)}{4 - 1}$

f(x) = x² -4x -3

f'(X) = 2x-4

f(1) = 1-4-3 = -6

f(4 ) = 16-16-3 = -3

then ,

⏩by mean value theorem,

$2x - 4 = \frac{ - 3 - ( - 6)}{4 - 1}$

$2x - 4 = \frac{3}{3}$

$2x - 4 = 1$

$x = \frac{5}{2}$

thus ,

c = 5/2 and 5/2 belongs to (1,4) such that

f'(c)

$= \frac{f(4) - f(1)}{4 - 1}$

⏩Hence ,

mean value theorem verified.

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