Question

find f such that f(x)=4x - 7, f(2)=0. f(x)=\square

Explanation:

Step1: Integrate \(f^{\prime}(x)\)

We know that if \(f^{\prime}(x)=4x - 7\), then by the power rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
\(\int(4x - 7)dx=\int4xdx-\int7dx\)
\(=4\times\frac{x^{2}}{2}-7x + C\)
\(=2x^{2}-7x + C\)

Step2: Use the condition \(f(2) = 0\) to find \(C\)

Substitute \(x = 2\) and \(f(2)=0\) into \(f(x)=2x^{2}-7x + C\)
\(0=2\times(2)^{2}-7\times2 + C\)
\(0=2\times4-14 + C\)
\(0 = 8-14 + C\)
\(0=-6 + C\), so \(C = 6\)

Answer:

\(f(x)=2x^{2}-7x + 6\)