Answer to Question #140985 in Calculus for Vanessa Samson

Question #140985
Find the function f whose tangent has slope (x^3-2/x^2+2) for each value of x and whose graph passes through the point ( 1, 3 ).
1
Expert's answer
2020-10-29T20:35:10-0400

The tangent slope is equal to the derivative, so for each value f(x)=x32x2+2f'(x) = x^3-\dfrac{2}{x^2}+2 . Therefore,

f(x)=(x32x2+2)dx=14x4+2x+2x+c.f(x) = \int (x^3-\dfrac{2}{x^2}+2)\,dx = \dfrac14 x^4+ \dfrac2x +2x + c.

Let us determine the constant c. We know that the graph of the function passes through the point (1, 3), so

3=1414+21+21+c    c=1.25.3 = \dfrac14\cdot1^4+ \dfrac21 +2\cdot1 + c \; \Rightarrow \; c = -1.25.

So f(x)==14x4+2x+2x1.25.f(x) == \dfrac14 x^4+ \dfrac2x +2x -1.25.


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