Problem

Find $f$ such that $f^{\prime}(x)=10 x-3, f(2)=0$ \[ f(x)= \]

Solution

Step 1 :The problem is asking for a function \(f(x)\) such that its derivative \(f'(x)\) is equal to \(10x - 3\) and \(f(2)\) is equal to 0. This is a problem of finding the antiderivative (or integral) of a function, and then adjusting the constant of integration such that the function passes through the point (2, 0).

Step 2 :The antiderivative of \(10x - 3\) is \(5x^2 - 3x + C\), where \(C\) is the constant of integration.

Step 3 :We can find the value of \(C\) by substituting \(x = 2\) and \(f(2) = 0\) into the equation and solving for \(C\).

Step 4 :Substituting these values into the equation gives us \(C + 14 = 0\), so \(C = -14\).

Step 5 :Substituting \(C = -14\) back into the equation gives us the function \(f(x) = 5x^2 - 3x - 14\).

Step 6 :\(\boxed{f(x) = 5x^2 - 3x - 14}\) is the function that satisfies the given conditions.

From Solvely APP
Source: https://solvelyapp.com/problems/8625/

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