What is the solution for f^'(x)=3x-4,f(-1)= 13/2 ?

The solution for f^'(x)=3x-4,f(-1)= 13/2 is f(x)=(3x^2)/2-4x+1

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f^'(x)=3x-4,f(-1)= 13/2

f^{^{'}} (x) = 3 x - 4, f (- 1) = \frac{13}{2}

f (x) = \frac{3 x ^{2}}{2} - 4 x + 1

Solution steps

f^{'} (x) = 3 x - 4

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

f (x) = \int 3 x - 4 d x

\int 3 x - 4 d x = \frac{3 x ^{2}}{2} - 4 x + c_{1}

f (x) = \frac{3 x ^{2}}{2} - 4 x + c_{1}

Apply initial conditions:

f (x) = \frac{3 x ^{2}}{2} - 4 x + 1

f (x) = \frac{3 x ^{2}}{2} - 4 x + 1

ln (196 sin^{2} (x))

\int \frac{1}{5 - 7 x} d x

y^{^{''''}} + 4 y^{^{''}} + 4 y = 0

\frac{\partial}{\partial x} (2 x y^{2} - 2 x^{3} y)

n = 0 \sum \infty \frac{( ( 1 ) ^{n} )}{n}

What is the solution for f^'(x)=3x-4,f(-1)= 13/2 ?
The solution for f^'(x)=3x-4,f(-1)= 13/2 is f(x)=(3x^2)/2-4x+1