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Q. 73

Expert-verifiedFound in: Page 198

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find a function that has the given derivative and value. In each case you can find the answer with an educated guess and check process it may be helpful to do some preliminary algebra

$f\text{'}\left(x\right)=3{x}^{5}-2{x}^{2}+4;f\left(0\right)=1$

The antiderivative can be given as $f\left(x\right)=\frac{{x}^{6}}{2}-\frac{2{x}^{3}}{3}+4x+1$

We are given the derivative as $f\text{'}\left(x\right)=3{x}^{5}-2{x}^{2}+4;f\left(0\right)=1$

We know that differentiating a power function decreases the power by one we can start with the function $f\left(x\right)={x}^{6}-{x}^{3}+4x+c$

On differentiating the above function we get,

$f\text{'}\left(x\right)=6{x}^{5}-3{x}^{2}+4$ which is nearly equal

Now we only adjust the coefficient

We get, $f\left(x\right)=\frac{{x}^{6}}{2}-\frac{2{x}^{3}}{3}+4x+c$

We are also given that $f\left(0\right)=1\phantom{\rule{0ex}{0ex}}Substittuingthisinthefunctionweget,\phantom{\rule{0ex}{0ex}}f\left(0\right)=c\phantom{\rule{0ex}{0ex}}c=1$

Hence the antiderivative becomes

$f\left(x\right)=\frac{{x}^{6}}{2}-\frac{2{x}^{3}}{3}+4x+1$

The antiderivative can be given as $f\left(x\right)=\frac{{x}^{6}}{2}-\frac{2{x}^{3}}{3}+4x+1$

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