Respuesta :
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) = [tex]$ \frac{f(b) - f(a)}{b - a} $[/tex]
Here, f(x) = x³ + x² - 1. a = 0, b =1
Since, f(x) is a polynomial, it is continuous and differentiable on the interval.
f'(x) = 3x² + 2x
⇒ f'(c) = 3c² + 2c
Using Mean value theorem, we have:
3c² + 2c = [tex]$ \frac{f(1) - f(0)}{1 - 0} $[/tex]
f(1) = 1 + 1 - 1 = 1
f(0) = 0 + 0 - 1 = - 1
[tex]$ \implies f'(c) = \frac{1 - (-1)}{1 - 0} $[/tex]
[tex]$ \implies f'(c) = \frac{2}{1} = 2 $[/tex]
Therefore, we have: 3c² + 2c = 2
Rearranging this, we have: 3c² + 2c - 2 = 0 which is a quadratic equation.
Now, we find the roots of the equation using the formula:
We have: c = [tex]$ \frac{- 2 \pm \sqrt{4 - 4(3)(2)}}{2.3} $[/tex]
= [tex]$ \frac{- 2 \pm \sqrt{4 + 24}}{6} $[/tex]
= [tex]$ \frac{- 2 \pm 2\sqrt{7}}{6} $[/tex]
= [tex]$ \frac{- 1 \pm \sqrt{7}}{3} $[/tex]
The roots are: c = [tex]$ \frac{- 1 + \sqrt{7}}{6} , \frac{- 1 - \sqrt{7}}{6} $[/tex]
Since, our root should lie between 0 and 1, we eliminate [tex]$ \frac{- 1 - \sqrt{7}}{6} $[/tex].
Hence, the value of c = [tex]$ \frac{- 1 + \sqrt{7}}{6} $[/tex]
So, we have proved the existence of 'c' and have determined the value of 'c' as well.
