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$f(x) = (2x+a)^3$, if $x <0$

$5bx+8$, if $0 ≤ x < 1$

$x^2 +12$, if $x ≥ 1$

I'm not quite show how to go about this question, I've tried placing 0 into the first equation as the left hand limit not sure if that is accurate.

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    you need for both formulas to have the same value at $x=1$. – 2017-01-30
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    Give your defintion for "continuous", please. – 2017-01-30

1 Answers 1

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For the function to be continuous, the limits from the left and right as $x \rightarrow 1$ must be the same, i.e.

$\begin{eqnarray} \lim_{x \rightarrow 1^-} f(x) & = & \lim_{x \rightarrow 1^+} f(x) \\ \lim_{x \rightarrow 1^-} \left( 5bx + 8 \right) & = & \lim_{x \rightarrow 1^+} \left( x^2 + 12 \right) \\ 5b \times 1 + 8 & = & 1^2 + 12 \\ 5b + 8 & = & 13 \\ \therefore b & = & 1 \end{eqnarray}$

Similarly, the left and right limits as $x \rightarrow 0$ must be the same, so you have:

$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x)$

which you can solve with a similar method, noting that you now have a value for $b$ that you can substitute in, letting you solve for $a$.