Question

1. find the value of the constant $b$ for which $f$ is continuous at $x = -1$ and determine $f(-1)$.

$f(x)=\begin{cases}x^{2}-5x & x < -1\bx^{3}-7 & xgeq -1end{cases}$

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = a\), \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). Here \(a=- 1\), so \(\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)\).

Step2: Calculate left - hand limit

For \(x
ightarrow - 1^{-}\), \(f(x)=x^{2}-5x\). Then \(\lim_{x
ightarrow - 1^{-}}f(x)=(-1)^{2}-5\times(-1)=1 + 5=6\).

Step3: Calculate right - hand limit

For \(x
ightarrow - 1^{+}\), \(f(x)=bx^{3}-7\). Then \(\lim_{x
ightarrow - 1^{+}}f(x)=b\times(-1)^{3}-7=-b - 7\).

Step4: Set left - hand and right - hand limits equal

Set \(-b - 7=6\). Add 7 to both sides: \(-b=6 + 7=13\), so \(b=-13\).

Step5: Find \(f(-1)\)

Since \(f(x)=bx^{3}-7\) for \(x\geq - 1\) and \(b = - 13\), then \(f(-1)=-13\times(-1)^{3}-7=13 - 7 = 6\).

Answer:

\(b=-13\), \(f(-1)=6\)