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# Find $f$.$f"(x) = 8x^3 + 5$, $\quad f(1) = 0$, $\quad f'(1) = 8$

## $f(x)=\frac{2}{5} x^{5}+\frac{5}{2} x^{2}+x-3.9$

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

given the second derivative, let's first find the first derivative. In other words, take the integral. So make the experiment one more derived by the new exponents. This is when C equals one. Now we're finding the integral again. So in other words, the original function D equals negative 3.9 There for the original function after box is to fifth xfff post five over too X squared plus X minus 3.9 or plus RC.

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