You need to notice that if the derivative is a linear function, then the primitive is a quadratic function, such that:
`f(x) = ax^2 + bx + c`
Differentiating f(x) yields:
`f'(x) = 2ax + b`
You need to set equal 2`x` and `2ax + b` , such that:
`2ax + b = 2x`
Comparing the coefficients of x both sides yields:
`2a = 2 => a = 1`
Notice that the free term b must be 0.
You should notice that c is not determined, yet. You may find out c using the information that the function passes through the point (1,0), such that:
`f(1) = 0 => f(1) = a*1^2 + b*1 + c `
`a + b + c = 0`
Replacing 1 for a and 0 for b yields:
`1 + 0 + c = 0 => c = -1`
Hence, evaluating the function under the given conditions, yields `f(x) = x^2 - 1.`
Therefore, if the point `(1,0)` lies on the function, then `f(x)=0` when `x=1` .
You can solve for C.
Thus, the complete function is `f(x)=x^2-1`