Math / Calculus

Question(1 point)
Let $f(x)=x+\sqrt{1-x}$ Find the local maximum and minimum values of $f$ using both the first and second derivative tests. Which method do you prefer? (That last question can be treated as rhetorical)
Below, type none if there are none. Points with local maximum values $\square$ Points with local minimum values $\square$

Studdy Solution

STEP 1

1. The function $f(x) = x + \sqrt{1-x}$ is defined for values of $x$ where the radicand $1-x$ is non-negative.
2. We will use both the first and second derivative tests to find local extrema.
3. The domain of $f(x)$ is $x \leq 1$ .

STEP 2

1. Determine the domain of the function.
2. Find the first derivative of the function.
3. Use the first derivative test to find critical points and determine if they are local maxima or minima.
4. Find the second derivative of the function.
5. Use the second derivative test to confirm the nature of the critical points.
6. Compare the results of both tests.

STEP 3

Determine the domain of the function $f(x) = x + \sqrt{1-x}$ .
The function is defined where the radicand is non-negative:
$1-x \geq 0$ $x \leq 1$
Thus, the domain of $f(x)$ is $(-\infty, 1]$ .

STEP 4

Find the first derivative of the function $f(x) = x + \sqrt{1-x}$ .
$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{1-x})$
$f'(x) = 1 - \frac{1}{2\sqrt{1-x}}$

STEP 5

Use the first derivative test to find critical points and determine if they are local maxima or minima.
Set the first derivative to zero to find critical points:
$1 - \frac{1}{2\sqrt{1-x}} = 0$
Solve for $x$ :
$\frac{1}{2\sqrt{1-x}} = 1$ $2\sqrt{1-x} = 1$ $\sqrt{1-x} = \frac{1}{2}$ $1-x = \frac{1}{4}$ $x = \frac{3}{4}$
Check the sign of $f'(x)$ around $x = \frac{3}{4}$ to determine the nature of the critical point:
- For $x < \frac{3}{4}$ , $f'(x) > 0$ . - For $x > \frac{3}{4}$ , $f'(x) < 0$ .
Thus, $x = \frac{3}{4}$ is a local maximum.

STEP 6

Find the second derivative of the function $f(x) = x + \sqrt{1-x}$ .
$f''(x) = \frac{d}{dx}\left(1 - \frac{1}{2\sqrt{1-x}}\right)$
$f''(x) = 0 + \frac{1}{4(1-x)^{3/2}}$

STEP 7

Use the second derivative test to confirm the nature of the critical point $x = \frac{3}{4}$ .
Evaluate $f''(x)$ at $x = \frac{3}{4}$ :
$f''\left(\frac{3}{4}\right) = \frac{1}{4(1-\frac{3}{4})^{3/2}} = \frac{1}{4(\frac{1}{4})^{3/2}} = \frac{1}{4(\frac{1}{8})} = 2$
Since $f''\left(\frac{3}{4}\right) > 0$ , the point $x = \frac{3}{4}$ is a local maximum.

STEP 8

Compare the results of both tests.
Both the first and second derivative tests confirm that $x = \frac{3}{4}$ is a local maximum.
Points with local maximum values: $x = \frac{3}{4}$
Points with local minimum values: $\text{none}$
The local maximum occurs at $x = \frac{3}{4}$ and there are no local minima.

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Studdy Solution

STEP 1

1. The function $f(x) = x + \sqrt{1-x}$ is defined for values of $x$ where the radicand $1-x$ is non-negative.
2. We will use both the first and second derivative tests to find local extrema.
3. The domain of $f(x)$ is $x \leq 1$ .

STEP 2

STEP 3

Determine the domain of the function $f(x) = x + \sqrt{1-x}$ .
The function is defined where the radicand is non-negative:
$1-x \geq 0$ $x \leq 1$
Thus, the domain of $f(x)$ is $(-\infty, 1]$ .

STEP 4

Find the first derivative of the function $f(x) = x + \sqrt{1-x}$ .
$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{1-x})$
$f'(x) = 1 - \frac{1}{2\sqrt{1-x}}$

STEP 5

STEP 6

Find the second derivative of the function $f(x) = x + \sqrt{1-x}$ .
$f''(x) = \frac{d}{dx}\left(1 - \frac{1}{2\sqrt{1-x}}\right)$
$f''(x) = 0 + \frac{1}{4(1-x)^{3/2}}$

STEP 7

STEP 8

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Studdy Solution

STEP 1

1. The function f(x)=x+1−x f(x) = x + \sqrt{1-x} f(x)=x+1−x​ is defined for values of x x x where the radicand 1−x 1-x 1−x is non-negative.2. We will use both the first and second derivative tests to find local extrema.3. The domain of f(x) f(x) f(x) is x≤1 x \leq 1 x≤1.

STEP 2

STEP 3

Determine the domain of the function f(x)=x+1−x f(x) = x + \sqrt{1-x} f(x)=x+1−x​.The function is defined where the radicand is non-negative:1−x≥0 1-x \geq 0 1−x≥0 x≤1 x \leq 1 x≤1Thus, the domain of f(x) f(x) f(x) is (−∞,1] (-\infty, 1] (−∞,1].

STEP 4

Find the first derivative of the function f(x)=x+1−x f(x) = x + \sqrt{1-x} f(x)=x+1−x​.f′(x)=ddx(x)+ddx(1−x) f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{1-x}) f′(x)=dxd​(x)+dxd​(1−x​)f′(x)=1−121−x f'(x) = 1 - \frac{1}{2\sqrt{1-x}} f′(x)=1−21−x​1​

STEP 5

STEP 6

STEP 7

STEP 8

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The function $f(x) = x + \sqrt{1-x}$ is defined for values of $x$ where the radicand $1-x$ is non-negative.
2. We will use both the first and second derivative tests to find local extrema.
3. The domain of $f(x)$ is $x \leq 1$ .

Determine the domain of the function $f(x) = x + \sqrt{1-x}$ .
The function is defined where the radicand is non-negative:
$1-x \geq 0$ $x \leq 1$
Thus, the domain of $f(x)$ is $(-\infty, 1]$ .

Find the first derivative of the function $f(x) = x + \sqrt{1-x}$ .
$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{1-x})$
$f'(x) = 1 - \frac{1}{2\sqrt{1-x}}$