Question

if we apply rolles theorem to the function $f(x) = 2x^2 - 8x - 1$ on the interval $0, 4$, how many values of $c$ exist such that $f(c) = 0$?

what is the value of $c$?

if we try to apply rolles thorem to the function $f(x) = 2x^2 - 8x - 1$ on the interval $-3, 11$, which of the following conditions is not met?

$\bigcirc$ continuty on $-3, 11$
$\bigcirc$ $f(a) = f(b)$
$\bigcirc$ differentiability on $-3, 11$

Explanation:

First Sub - Question: How many values of \( c \) exist such that \( f^{\prime}(c)=0 \)?

Step 1: Check Rolle's Theorem conditions

Rolle's Theorem states that if a function \( y = f(x) \) is continuous on \([a,b]\), differentiable on \((a,b)\), and \( f(a)=f(b) \), then there exists at least one \( c\in(a,b) \) such that \( f^{\prime}(c) = 0 \).

First, check \( f(0) \) and \( f(4) \) for \( f(x)=2x^{2}-8x - 1 \):

\( f(0)=2(0)^{2}-8(0)-1=- 1 \)

\( f(4)=2(4)^{2}-8(4)-1=2\times16 - 32 - 1=32 - 32 - 1=-1 \)

Since \( f(0) = f(4)\), and \( f(x)=2x^{2}-8x - 1 \) is a polynomial (so it is continuous on \([0,4]\) and differentiable on \((0,4)\)).

Step 2: Find the derivative of \( f(x) \)

The derivative of \( f(x)=2x^{2}-8x - 1 \) using the power rule \( (x^{n})^\prime=nx^{n - 1} \) is:

\( f^{\prime}(x)=4x-8 \)

Step 3: Solve \( f^{\prime}(c)=0 \)

Set \( f^{\prime}(c)=0 \), so \( 4c - 8=0 \)

Add 8 to both sides: \( 4c=8 \)

Divide both sides by 4: \( c = 2 \)

Since the derivative is a linear function (degree 1), there is only 1 solution for \( c \) in the interval \((0,4)\).

Step 1: Use the derivative from the first sub - question

We found that \( f^{\prime}(x)=4x - 8 \)

Step 2: Solve \( f^{\prime}(c)=0 \)

Set \( 4c-8 = 0 \)

\( 4c=8 \)

\( c=\frac{8}{4}=2 \)

Step 1: Check continuity

\( f(x)=2x^{2}-8x - 1 \) is a polynomial, so it is continuous on \([-3,11]\).

Step 2: Check differentiability

Polynomial functions are differentiable everywhere, so \( f(x) \) is differentiable on \((-3,11)\) (and thus on \([-3,11]\) in the sense of differentiability on the open interval which is required for Rolle's Theorem, but even as a polynomial it is differentiable on the closed interval too).

Step 3: Check \( f(a)=f(b) \)

Calculate \( f(-3) \) and \( f(11) \):

\( f(-3)=2(-3)^{2}-8(-3)-1=2\times9 + 24-1=18 + 24-1=41 \)

\( f(11)=2(11)^{2}-8(11)-1=2\times121-88 - 1=242-88 - 1=153 \)

Since \( f(-3)
eq f(11) \), the condition \( f(a) = f(b) \) is not met.

Answer:

1

Second Sub - Question: What is the value of \( c \)?