Comparing f(x) and g(x) for All x in R

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The discussion centers on comparing the functions f(x) = 5(√(x^2 +1)) and g(x) = 3x + 4. The minimum point of f(x) is found to be greater than g(x), and both functions are tangent at the point (3/4). To demonstrate that f(x) is greater than or equal to g(x) for all x in R, the minimum of the difference function f(x) - g(x) is analyzed, revealing it equals zero at the tangential point. The conclusion confirms that f(x) is indeed greater than or equal to g(x) across the entire domain. This establishes the relationship between the two functions effectively.
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Homework Statement


the two functions f(x) = 5(√(x^2 +1)) g(x) = 3x + 4.

Homework Equations

The Attempt at a Solution


I can get the minimum point of f(x) and it is bigger than g(x) and that point, however g(x) is tangential to the curve f(x) at point 3/4.
what else do i miss to show that f(x) is bigger or equal than g(x) for all x in R?
 
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Dank2 said:

Homework Statement


the two functions f(x) = 5(√(x^2 +1)) g(x) = 3x + 4.

Homework Equations

The Attempt at a Solution


I can get the minimum point of f(x) and it is bigger than g(x) and that point, however g(x) is tangential to the curve f(x) at point 3/4.
what else do i miss to show that f(x) is bigger or equal than g(x) for all x in R?

Find the minimum of the difference function ##f(x) - g(x) = 5 \sqrt{x^2+1} -(3x+4)##.

BTW: the word is parabola, not parabula; and anyway, you do not have a parabola anywhere in this problem.
 
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Ray Vickson said:
Find the minimum of the difference function ##f(x) - g(x) = 5 \sqrt{x^2+1} -(3x+4)##.

BTW: the word is parabola, not parabula; and anyway, you do not have a parabola anywhere in this problem.
thanks

its point 3/4. and it is the absolute minimum of the graph that's equal to 0, therefore f(x) >= g(x).
 
Dank2 said:
thanks

its point 3/4. and it is the absolute minimum of the graph that's equal to 0, therefore f(x) >= g(x).
Right!
 
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