Question

Question #96b06

Answer 1

See below.

Explanation:

`e^(-x)` is monotonic strictly decreasing for `x in [0, oo)` and
`x^3` is monotonic strictly increasing for `x in [0,1)` both functions are continuous so they must cross for a point `x_0` such that
`e^(-x_0) = x_0^3 < 1`. Note that for `x in [0,1)` we have
`x_0^3 < 1` and `e^(-x_0) < 1`

Answer 2

See answer below

Explanation:

You must use the intermediate value theorem
let `f(x)=e^-x-x^3` this is a continuous function on the interval `(0,1)`
`f(0)=e^0-0=1`
and `f(1)=e^-1-1<0`
Then there is a value `c ∈ (0,1)` such that `f(1 )< f(c) < f(0)`

Answer 3

It is explained below

Explanation:

Given `e^(-x)=x^3`, take natural log on both sides. It would be `-xln e=3ln x`
Or -x = 3 ln x. This expression signifies that for all real x, x cannot be less than or equal to 0.

Now, in the logarithmic equation the LHS has - x, while it has a positive expression 3ln x on the RHS. This would be possible only if x is less than 1

Thus it is proved that x would lie between 0 and 1