Question

Given an interval $[a, b]$ that satisfies hypothesis of Rolle's thorem for the function

$f(x) = x^4 + x^2 - 2$. It is known that $a = -1$. Find the value of $b$.

Answer 1

1

Answer 2

The problem asks for the value of $b$ such that the function $f(x) = x^4 + x^2 - 2$ satisfies the hypotheses of Rolle's theorem on the interval $[a, b]$, given that $a = -1$.

Rolle's theorem states that if a function $f(x)$ is continuous on the closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f'(c) = 0$.

The given function is $f(x) = x^4 + x^2 - 2$, which is a polynomial function.

1. Polynomial functions are continuous everywhere, so $f(x)$ is continuous on any closed interval $[a, b]$.
2. Polynomial functions are differentiable everywhere, so $f(x)$ is differentiable on any open interval $(a, b)$.

For Rolle's theorem hypotheses to be satisfied, the third condition $f(a) = f(b)$ must also hold. We are given $a = -1$. So, we must have $f(-1) = f(b)$.

First, calculate $f(-1)$:

$f(-1) = (-1)^4 + (-1)^2 - 2 = 1 + 1 - 2 = 0$.

Now, we need to find the value(s) of $b$ such that $f(b) = 0$.

$f(b) = b^4 + b^2 - 2$.

Set $f(b) = 0$:

$b^4 + b^2 - 2 = 0$.

This is a quadratic equation in terms of $b^2$. Let $y = b^2$. Then the equation becomes:

$y^2 + y - 2 = 0$.

We can solve this quadratic equation by factoring:

$(y + 2)(y - 1) = 0$.

This gives two possible values for $y$:

$y + 2 = 0 \implies y = -2$.

$y - 1 = 0 \implies y = 1$.

Substitute back $y = b^2$:

Case 1: $b^2 = -2$. This equation has no real solutions for $b$, since the square of a real number cannot be negative.

Case 2: $b^2 = 1$. This equation has two real solutions for $b$: $b = \sqrt{1} = 1$ or $b = -\sqrt{1} = -1$.

So, the possible real values for $b$ such that $f(b) = f(-1) = 0$ are $1$ and $-1$.

The interval is given as $[a, b] = [-1, b]$. For Rolle's theorem to apply in the usual sense (guaranteeing a $c$ in the open interval $(a, b)$), we typically require $a < b$. Given $a = -1$, we need $b > -1$.

From the possible values $b = 1$ and $b = -1$, only $b = 1$ satisfies the condition $b > -1$. If $b = -1$, the interval is $[-1, -1]$, which is a single point. While continuity and differentiability hold, the open interval $(-1, -1)$ is empty, making the conclusion of Rolle's theorem (existence of $c \in (a, b)$) trivially true but not useful in finding a critical point in an open interval. Standard applications of Rolle's theorem assume $a < b$.

Therefore, the value of $b$ that forms a standard interval $[a, b]$ with $a = -1$ satisfying the hypotheses of Rolle's theorem is $b = 1$. The interval is $[-1, 1]$.

Let's verify the hypotheses for $[-1, 1]$:

1. $f(x) = x^4 + x^2 - 2$ is continuous on $[-1, 1]$ (as it is a polynomial).
2. $f(x) = x^4 + x^2 - 2$ is differentiable on $(-1, 1)$ (as it is a polynomial).
3. $f(-1) = 0$ and $f(1) = 1^4 + 1^2 - 2 = 1 + 1 - 2 = 0$. So, $f(-1) = f(1)$.

All hypotheses are satisfied for the interval $[-1, 1]$.

The value of $b$ is $1$.