The tangent slope is equal to the derivative, so for each value f′(x)=x3−x22+2 . Therefore,
f(x)=∫(x3−x22+2)dx=41x4+x2+2x+c.
Let us determine the constant c. We know that the graph of the function passes through the point (1, 3), so
3=41⋅14+12+2⋅1+c⇒c=−1.25.
So f(x)==41x4+x2+2x−1.25.
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