Question

show that root of equation Xe^x =1 lies between 0 and 1​

Answer 1

Step-by-step explanation:

X.e^x = 1

X = e^-x

then X = + Infinity

it means there is no roots between 0,1

Answer 2

To Prove:

That the equation xeˣ = 1 has a root between 0 and 1.

Proof:

Let us assume y = xeˣ - 1 = f(x) i.e., y is a function of 'x'.

Now, we know that for a strictly increasing or decreasing function, a root lies between 'a' and 'b' only if f(a)f(b) < 0 i.e., f(a) and f(b) are of different signs.

Here, let us check if our function f(x) = xeˣ - 1 is a monotonic function or not.

So, To check the monotonicity of a function, we check it by calculating its differentiation.

∴ f(x) = (x + 1)eˣ

∴ f'(x)  ≥ 0 for x ≥ -1

This proves that our function is strictly increasing in the region where we are going to check its root,

Now, f(0) × f(1) = (-1) × (e - 1)

i.e., f(0)f(1) = (1 - e) < 0

Hence, it is proved that the function f(x) has its root between 0 and 1.

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