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**DarkestEvil** i cant seem to understand this concept, its very confusing, can anyone give me any tips?

4. Suppose that the function $\displaystyle f$ has a continuous second derivative for all $\displaystyle x$, and that $\displaystyle f(0)=2$, $\displaystyle f'(o)=-3$, and $\displaystyle f''(0)=0$. Let $\displaystyle g$ be a function whose derivative is given by $\displaystyle g'(x)=e^{-2x} (3f(x)+2f'(x))$ for all $\displaystyle x$.

a. Write an equation of the line tangent to the graph of the $\displaystyle f$ at the point where $\displaystyle x=0$

they gave you everything you need to do that ... f(0) = 2 and f'(0) = -3

b. Is there sufficient information to determine whether or not the graph of $\displaystyle f$ has a point of inflection when $\displaystyle x=0$? Explain your answer.

what do you need to tell you that f(x) has inflection point(s) ?

c. Given that $\displaystyle g(0)=4$, write an equation of the line tangent to the graph of $\displaystyle g$ at the point where $\displaystyle x=0$

another tangent line problem ... they gave you the point, g(0) = 4 , all you need is the value of g'(0) for the slope of the line ... can you find that?

d. Show that $\displaystyle g''(x)=e^{-2x} (-6f(x)-f'(x)+2f''(x))$. Does $\displaystyle g$ have the local maximum at $\displaystyle x=0$? Justify your answer.

fairly straightforward ... they gave you g'(x) and want you to find g''(x) simplified to the given form.