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Oct 27, 2021 · f(x) = x - 5, find f(2): Basically substitute x with any value then solve the equation. f(2) = 2 - 5 = -3. Other examples: f(x) = x^2, find f(2): f(2) = 2^2 = 4. f(x) = 2x + 3, find f(5): f(5) …

Write f (x) = (x−5)2 f (x) = (x - 5) 2 as an equation. Interchange the variables. Solve for y y. Tap for more steps... Replace y y with f −1(x) f - 1 (x) to show the final answer. Verify if f −1(x) = √x+ …

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f (x) = 5x2 −5 f (x) = 5 x 2 - 5. Plug in the components. f (x+h)−f (x) h = 5h2 +10hx+5x2 −5−(5x2 −5) h f (x + h) - f (x) h = 5 h 2 + 10 h x + 5 x 2 - 5 - (5 x 2 - 5) h. Simplify. Tap for more steps...

Jun 19, 2016 · So if you have #f(2)# you're applying the function with your value of #x# as 2, so we simply substitute 2 into the equation where there is an #x# so in this case: #f(x) = 3x - 5# #f(2) …

For f(x) = 5x - x^2, use the definition of a derivative to find f'(x) Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.

Write f (x) = (x−5)2 f (x) = (x - 5) 2 as an equation. Use the vertex form, y = a(x−h)2 +k y = a (x - h) 2 + k, to determine the values of a a, h h, and k k. Since the value of a a is positive, the …

How do you find local maximum value of f using the first and second derivative tests: \displaystyle{f{{\left({x}\right)}}}={5}{x}-{3} ?

The domain and range of the function f: R → R defined by: f = {(x+1, x+5): x ∈ {0, 1, 2, 3, 4, 5}}

Sep 21, 2019 · If f (x) = 6^ (x − 2), find f (5). For the function f (x), the expression f (5) just means plug 5 into the equation rule associated with the function. The function rule associated with f …

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If F(x) = x^2 − 5x, find F '(5) and use it to find an equation of the tangent line to the curve y = x^2 − 5x at the point (5, 0). Here’s the best way to solve it. Solution

Oct 19, 2023 · Step-by-step Solved, Expert Educator: If f (x) = 5* , find f (2) and f ( - 2). VIDEO ANSWER: The function has to be evaluated at the values. We replace X with zero and then …

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x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} …

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} …

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