Using the Intermediate Value Theorem, verify (show) that f(x) = x3-100x+2 has or does not have a zero on the interval (0.01, 0.1).

Here's one way to do it.

f

(

x

)

=

x

3

−

2

x

2

+

3

x

−

5

Evaluate

f

(

1

)

and

f

(

2

)

and note that the function is continuous with values on both sides of

0

Explanation:

Note:

x

3

−

2

x

2

+

3

x

=

5

is a polynomial equation; it is not a polynomial function. Reference to an equation having a zero in an interval is meaningless.

f

(

x

)

=

x

3

−

2

x

2

+

3

x

does not have a zero in the interval

[

1

,

2

]

That only leaves

f

(

x

)

=

x

3

−

2

x

+

3

x

−

5

as the intended function

f

(

1

)

=

−

3

f

(

2

)

=

+

1

Since

f

(

x

)

is continuous in the interval

[

1

,

2

]

it must have have values for everything in the range

[

−

3

,

+

1

]

including zero.

Step-by-step explanation:

the answer is b.)

< 1=< 2

income. 10x+1300

expenses 3x+(1800+1500)

expenses 3x+3300

to break even:

income = expense,

10x + 1300 = 3x + 3300,

solve for x.

step-by-step explanation:

mark brainliest