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Find f such that f'(x)= x² - 6 and f(0) = 1.

User Mrr
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1 Answer

2 votes

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

f'(x)= x² - 6

f(0) = 1

f(x) = ?

Step 02:


\int (x^2-6)dx\text{ = }(x^3)/(3)-6x
\begin{gathered} f(0)=(0^3)/(3)-6\cdot0=(0)/(3)^{}-0=0 \\ f(x)\text{ =}(x^3)/(3)-6x+1 \end{gathered}

The answer is:

f(x) = x^3 / 3 - 6x + 1

User Deadpool
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