Mathematics, 20.12.2020 18:00, sanociahnoel

What is the largest possible integral value in the domain of the real-valued function f(x)= $\frac{1}{\sqrt{800-2x} }$

Answers: 3

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answered: Lesquirrel

Max Value: x = 400

General Formulas and Concepts:

Algebra I

Domain is the set of x-values that can be inputted into function f(x)

Calculus

AntiderivativesIntegral Property: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$ Integration Method: U-Substitution[Integration] Reverse Power Rule: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C$

Step-by-step explanation:

Step 1: Define

$f(x) = \frac{1}{\sqrt{800-2x} }$

Step 2: Identify Variables

Using U-Substitution, we set variables in order to integrate.

$u = 800-2x\\du = -2dx$

Step 3: Integrate

Define: $\int {f(x)} \, dx$ Substitute: $\int {\frac{1}{\sqrt{800-2x} } } \, dx$ [Integral] Int Property: $-\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx$ [Integral] U-Sub: $-\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du$ [Integral] Rewrite: $-\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du$ [Integral - Evaluate] Reverse Power Rule: $-\frac{1}{2}(2\sqrt{u}) + C$ Simplify: $-\sqrt{u} + C$ Back-Substitute: $-\sqrt{800-2x} + C$ Factor: $-\sqrt{-2(x - 400)} + C$

Step 4: Identify Domain

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

answered: Guest

Hope i can you!

answered: Guest

1) is right

step-by-step explanation:

Can someone helm me with this problem plz and show work

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What is the largest possible integral value in the domain of the real-valued function f(x)=...