Answer:
Use Mean Value theorem.
Step-by-step explanation:
Statement: If f(x) is continuous on [a, b] and differentiable on (a, b) then there is at least one 'c' (a < c < b), then we have:
f'(c) = 
Here, f(x) = x³ + x² - 1. a = 0, b =1
Since, f(x) is a polynomial, it is continuous and differentiable on the interval.
f'(x) = 3x² + 2x
⇒ f'(c) = 3c² + 2c
Using Mean value theorem, we have:
3c² + 2c = 
f(1) = 1 + 1 - 1 = 1
f(0) = 0 + 0 - 1 = - 1


Therefore, we have: 3c² + 2c = 2
Rearranging this, we have: 3c² + 2c - 2 = 0 which is a quadratic equation.
Now, we find the roots of the equation using the formula:
We have: c = 
= 
= 
= 
The roots are: c = 
Since, our root should lie between 0 and 1, we eliminate
.
Hence, the value of c = 
So, we have proved the existence of 'c' and have determined the value of 'c' as well.