Question

Show there is a number c ,with 0<_c<_1,such that f(c)=0 for the equation f(x)=x^3+x^2-1

Answer 1

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) = $\frac{f(b) - f(a)}{b - a}$

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 = $\frac{f(1) - f(0)}{1 - 0}$

f(1) = 1 + 1 - 1 = 1

f(0) = 0 + 0 - 1 = - 1

$\implies f'(c) = \frac{1 - (-1)}{1 - 0}$

$\implies f'(c) = \frac{2}{1} = 2$

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 = $\frac{- 2 \pm \sqrt{4 - 4(3)(2)}}{2.3}$

= $\frac{- 2 \pm \sqrt{4 + 24}}{6}$

= $\frac{- 2 \pm 2\sqrt{7}}{6}$

= $\frac{- 1 \pm \sqrt{7}}{3}$

The roots are: c = $\frac{- 1 + \sqrt{7}}{6} , \frac{- 1 - \sqrt{7}}{6}$

Since, our root should lie between 0 and 1, we eliminate $\frac{- 1 - \sqrt{7}}{6}$.

Hence, the value of c = $\frac{- 1 + \sqrt{7}}{6}$

So, we have proved the existence of 'c' and have determined the value of 'c' as well.