Question

Roots of equation help. *Question attached below* Will give medal

Answer 1

So I attempted part a to this using the Intermediate Value Theorem. Is it correct if I let k=1 then showed the root existed in the interval?
As for part b, I am quite clueless.

Answer 2

sorry idk this :(

Answer 3

if you plug in x = 0 and x = 1 , what do you get

Answer 4

Let f(x) = x^3 + kx -1

f(0) = 0^3 + k*0 -1 = -1
f(1) = 1^3 + k*1 -1 = k + 0 = k
but we assumed that k > 0

So far we have f(0) < 0 , and f(1) > 0.

Since f is continuous function (since it is a polynomial), by intermediate value theorem there is some number c such that f(c) = 0 for c between 0 and 1

Answer 5

I'm not allowed to help you on a test

Answer 6

Ahhh, now I see how to do it without letting k be any set value. Thank you! @perl

Answer 7

right, since f(1) = k , and we assumed k>0 in the directions
f(1) > 0

Answer 8

Do I use a graph for part b? To show that it only has one root?

Answer 9

you can use an argument from contradiction

Answer 10

since the derivative of f(x) is
f '(x) = 3x^2 + k , and we assumed k > 0

then f(x) is always increasing.

Answer 11

Thank you!