Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem.

Question

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. (Enter your answers as a comma-separated list.)

f(x)=(sqrt of x)-1/5x (0,25)

Answer 1

Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(c) = 0 for some c with a ≤ x ≤ b.

f(x) is for sure continuous on [0,25] √

f'(x) is for sure differentiable on (0,25) √

f(a)=f(b) because f(0)=0=f(25) √

f(x) = √x - (1/5)x

f'(x) = 1/(2√x) - 1/5

f'(c) = 1/(2√c) - 1/5

0 = 1/(2√c) - 1/5

1/5 = 1/(2√c)

2√c/5 = 1

√c = 5/2

c = 25/4

Thus, the value of c=25/4 satisfies the conclusion of Rolle's Theorem.

Answer 2

f(0) = sqr(0) -1/5(0) = undefined

f(25) = sqr25 - 1/5(25) = 5-1/125 = 4.992

f(0) doesn't equal f(25)

it doesn't satisfy one requirement

no need to check the other 2 requirements

of continuity and differentiability

over the close&open interval

BUT you more likely,, almost certainly meant to write

f(x) = sqrx -(1/5)x

then

f(0) = 0

f(25) = sqr5 -25/5 = 0

satisfying one requirement of Rolles' Theorem

then check the differentiability of the function over the open interval (0,25)

and the continuity over the closed interval [0,25]

f'(x) = (1/2)/sqrx -1/5 which is differentiable except at x=0, but that's not in the Open interval

check the continuity by graphing f(x)

it's half a parabola, the top half of a rightward opening parabola with vertex at the origin

but the half parabola is shifted down by x/5

try a graphing calculator to see if it's a continuous graph with no "holes" or discontinuity

if you can draw the graph with a pencil without lifting up your pencil it's continuous

f'(c) = [f(25)- f(0)]/(25-0) = 0/25 = 0

= (1/2)/sqrx - 1/5 =0

1/2sqrx = 1/5

2sqrx = 5

sqrx = 5/2

x = (5/2)^2 = 25/4

f'(6.25) = f'(25/4) = (1/2)/sqr(25/4) - 1/5 =.5/sqr6.25 - .2 = .5/.25 -.2 = .2-.2=0

f(25/4) = sqr(25/4) -1/5 = 5/2 -1/5 = 2.5-.2 = 2.3

slope of the tangent line to the curve at the point (6.25,2.3) = the slope of the line

y = 0 which intersects the curve at the origin (0,0) and at (25,0)

0 < 6.25 < 25

c=6.25