Continuity by Definition – Continuity check by definition to a function with parameters – Exercise 898

Exercise

Given the function (a,b and c parameters)

f(x)={x3+ax2+bx+c(x+1)2,x13,x=1f(x) = \begin{cases} \frac{x^3+ax^2+bx+c}{{(x+1)}^2}, &\quad x \neq -1\\ -3, &\quad x = -1\\ \end{cases}

For which values of the parameters is the function continuous?

Final Answer


a=0,b=3,c=2a=0, b=-3, c=-2

Solution

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