1. ## Continuity of Functions

Hey everyone! I have these two online problems that I cannot figure out for the life of me. Please help!! I understand continuity but these are terrible.

1.) The function f is given by the formula
when and by the formula

when .

What value must be chosen for a in order to make this function continuous at -4?

2.) For what value of the constant c is the function f continuous on where

Thank you so much guys!

2. For the first one...

$\displaystyle f(x)=\frac{4x^3+15x^2+4x+32}{x+4} \iff \frac{(x+4)(4x^2-x+8)}{(x+4)}=(4x^2-x+8)$

so evaluate the above expression at -4 and set equal to
$\displaystyle 3x^2-2x+a$ evaluated at -4 and solve for a.

Do same thing for part 2 except use x=5.

Good luck.

3. Hello, funnylookinkid!

1) The function $\displaystyle f(x)$ is given by: .$\displaystyle f(x) \;=\;\begin{Bmatrix}\dfrac{4x^3+15x^2+4x+32}{x+4} & x < -4 \\ \\[-1mm] 3x^2 - 2x + a & x \geq -4\end{Bmatrix}$

What value must be chosen for $\displaystyle a$ in order to make this function continuous at -4?

For $\displaystyle x < -4$ the first function is: .$\displaystyle \frac{(x+4)(4x^2-x+8)}{x+4} \;=\;4x^2 - x + 8$

So we have: .$\displaystyle f(x)\;=\;\begin{Bmatrix}4x^2 - x + 8 & x < -4 \\ 3x^2 - 2x + a & x \geq -4 \end{Bmatrix}$

To be continuous at $\displaystyle x = -4$, the two branches must be equal.

. . $\displaystyle 4(\text{-}4)^2 - (\text{-}4 ) + 8 \;=\;3(\text{-}4)^2 - 2(\text{-}4) + a \quad\Rightarrow\quad\boxed{ a \:=\:20}$

2) For what value of the constant $\displaystyle c$ is the function $\displaystyle f(x)$ continuous?

. . $\displaystyle f(x) \;=\;\begin{Bmatrix}cx + 4 & x \leq 5 \\ cx^2-4 & x > 5 \end{Bmatrix}$

To be continuous at $\displaystyle x = 5$, the two branches must be equal.

. . $\displaystyle 5x + 4 \:=\:25x - 4\quad\Rightarrow\quad\boxed{ c \:=\:\frac{2}{5}}$

4. Thank you guys so much! The problem was in me not knowing how to factor the first and me not opening my eyes for the second, haha!