Math / Calculus

QuestionDetermine whether Rolle's Theorem can be applied to the function on the given interval; if so, find the value(s) of $c$ guaranteed by the theorem. (Enter your answers as a comma-separated list. If Rolle's Theorem does not apply, enter DNE.) $f(x)=\sqrt{x}(2-x) \text { on }[0,2]$ $c=$ Tutorial

Studdy Solution

STEP 1

What is this asking? Can we use Rolle's Theorem on this square root function, and if so, what are the c values it guarantees? Watch out! Rolle's Theorem has specific conditions, so we need to make sure they're met before applying it!

STEP 2

1. Check Continuity
2. Check Differentiability
3. Check Endpoint Equality
4. Find the Derivative
5. Solve for c

STEP 3

Let's check if $f(x) = \sqrt{x}(2-x)$ is continuous on the closed interval $[0, 2]$ .
The square root function $\sqrt{x}$ is continuous for $x \ge 0$ , and $(2-x)$ is a polynomial, so it's continuous everywhere.
Their product is continuous wherever both factors are defined, which includes the interval $[0, 2]$ .
So, $f(x)$ is indeed continuous on $[0, 2]$ !

STEP 4

Now, let's see if $f(x)$ is differentiable on the open interval $(0, 2)$ .
We can rewrite $f(x)$ as $2\sqrt{x} - x\sqrt{x} = 2x^{\frac{1}{2}} - x^{\frac{3}{2}}$ .

STEP 5

Taking the derivative, we get $f'(x) = 2 \cdot \frac{1}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = \frac{1}{\sqrt{x}} - \frac{3\sqrt{x}}{2}$ .

STEP 6

$f'(x)$ is defined for all $x > 0$ , so $f(x)$ is differentiable on $(0, 2)$ .
Great!

STEP 7

Let's check if $f(0)$ equals $f(2)$ .
We have $f(0) = \sqrt{0}(2-0) = 0$ and $f(2) = \sqrt{2}(2-2) = 0$ .
Since $f(0) = f(2) = 0$ , this condition is satisfied!

STEP 8

We already found the derivative in a previous step: $f'(x) = \frac{1}{\sqrt{x}} - \frac{3\sqrt{x}}{2}$ .
Now we're ready to use it!

STEP 9

Rolle's Theorem says there's at least one $c$ in $(0, 2)$ where $f'(c) = 0$ .
So, let's set $f'(c) = 0$ and solve for $c$ : $0 = \frac{1}{\sqrt{c}} - \frac{3\sqrt{c}}{2}$ $\frac{1}{\sqrt{c}} = \frac{3\sqrt{c}}{2}$ Multiplying both sides by $2\sqrt{c}$ gives us: $2\sqrt{c} \cdot \frac{1}{\sqrt{c}} = 2\sqrt{c} \cdot \frac{3\sqrt{c}}{2}$ $2 = 3c$ $c = \frac{2}{3}$ Since $\frac{2}{3}$ is in the interval $(0, 2)$ , we found our $c$ value!

STEP 10

$c = \frac{2}{3}$

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Studdy Solution

STEP 1

What is this asking? Can we use Rolle's Theorem on this square root function, and if so, what are the c values it guarantees? Watch out! Rolle's Theorem has specific conditions, so we need to make sure they're met before applying it!

STEP 2

1. Check Continuity
2. Check Differentiability
3. Check Endpoint Equality
4. Find the Derivative
5. Solve for c

STEP 3

STEP 4

Now, let's see if $f(x)$ is differentiable on the open interval $(0, 2)$ .
We can rewrite $f(x)$ as $2\sqrt{x} - x\sqrt{x} = 2x^{\frac{1}{2}} - x^{\frac{3}{2}}$ .

STEP 5

Taking the derivative, we get $f'(x) = 2 \cdot \frac{1}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = \frac{1}{\sqrt{x}} - \frac{3\sqrt{x}}{2}$ .

STEP 6

$f'(x)$ is defined for all $x > 0$ , so $f(x)$ is differentiable on $(0, 2)$ .
Great!

STEP 7

Let's check if $f(0)$ equals $f(2)$ .
We have $f(0) = \sqrt{0}(2-0) = 0$ and $f(2) = \sqrt{2}(2-2) = 0$ .
Since $f(0) = f(2) = 0$ , this condition is satisfied!

STEP 8

We already found the derivative in a previous step: $f'(x) = \frac{1}{\sqrt{x}} - \frac{3\sqrt{x}}{2}$ .
Now we're ready to use it!

STEP 9

STEP 10

$c = \frac{2}{3}$

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Studdy Solution

STEP 1

What is this asking? Can we use Rolle's Theorem on this square root function, and if so, what are the *c* values it guarantees? Watch out! Rolle's Theorem has specific conditions, so we need to make sure they're met before applying it!

STEP 2

1. Check Continuity2. Check Differentiability3. Check Endpoint Equality4. Find the Derivative5. Solve for *c*

STEP 3

STEP 4

Now, let's **see if** f(x)f(x)f(x) **is differentiable** on the open interval (0,2)(0, 2)(0,2).We can rewrite f(x)f(x)f(x) as 2x−xx=2x12−x322\sqrt{x} - x\sqrt{x} = 2x^{\frac{1}{2}} - x^{\frac{3}{2}}2x​−xx​=2x21​−x23​.

STEP 5

STEP 6

f′(x)f'(x)f′(x) is defined for all x>0x > 0x>0, so f(x)f(x)f(x) is differentiable on (0,2)(0, 2)(0,2).Great!

STEP 7

Let's **check if** f(0)f(0)f(0) **equals** f(2)f(2)f(2).We have f(0)=0(2−0)=0f(0) = \sqrt{0}(2-0) = 0f(0)=0​(2−0)=0 and f(2)=2(2−2)=0f(2) = \sqrt{2}(2-2) = 0f(2)=2​(2−2)=0.Since f(0)=f(2)=0f(0) = f(2) = 0f(0)=f(2)=0, this condition is satisfied!

STEP 8

We already **found the derivative** in a previous step: f′(x)=1x−3x2f'(x) = \frac{1}{\sqrt{x}} - \frac{3\sqrt{x}}{2}f′(x)=x​1​−23x​​.Now we're ready to use it!

STEP 9

STEP 10

c=23c = \frac{2}{3}c=32​

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What is this asking? Can we use Rolle's Theorem on this square root function, and if so, what are the c values it guarantees? Watch out! Rolle's Theorem has specific conditions, so we need to make sure they're met before applying it!

1. Check Continuity
2. Check Differentiability
3. Check Endpoint Equality
4. Find the Derivative
5. Solve for c

Now, let's see if $f(x)$ is differentiable on the open interval $(0, 2)$ .
We can rewrite $f(x)$ as $2\sqrt{x} - x\sqrt{x} = 2x^{\frac{1}{2}} - x^{\frac{3}{2}}$ .

$f'(x)$ is defined for all $x > 0$ , so $f(x)$ is differentiable on $(0, 2)$ .
Great!

Let's check if $f(0)$ equals $f(2)$ .
We have $f(0) = \sqrt{0}(2-0) = 0$ and $f(2) = \sqrt{2}(2-2) = 0$ .
Since $f(0) = f(2) = 0$ , this condition is satisfied!

We already found the derivative in a previous step: $f'(x) = \frac{1}{\sqrt{x}} - \frac{3\sqrt{x}}{2}$ .
Now we're ready to use it!

$c = \frac{2}{3}$