Question

find the value of the constant a that makes the following function continuous on (-∞, ∞).
f(x)=\begin{cases}\frac{2x^{3}-x^{2}+7x - 8}{x - 1}&\text{if }x<1\\-5x^{2}-3x + a&\text{if }xgeq1end{cases}
a = (square)

Explanation:

Step1: Simplify the left - hand limit

For \(x < 1\), simplify \(\frac{2x^{3}-x^{2}+7x - 8}{x - 1}\). Use polynomial long - division or factor by grouping.
We know that \(2x^{3}-x^{2}+7x - 8=2x^{3}-2x^{2}+x^{2}-x + 8x-8=2x^{2}(x - 1)+x(x - 1)+8(x - 1)=(x - 1)(2x^{2}+x + 8)\). So \(\lim_{x
ightarrow1^{-}}\frac{2x^{3}-x^{2}+7x - 8}{x - 1}=\lim_{x
ightarrow1^{-}}(2x^{2}+x + 8)\).
Substitute \(x = 1\) into \(2x^{2}+x + 8\): \(2(1)^{2}+1 + 8=2 + 1+8 = 11\).

Step2: Calculate the right - hand limit

For \(x\geq1\), \(f(x)=-5x^{2}-3x + a\). Then \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(-5x^{2}-3x + a)=-5(1)^{2}-3(1)+a=-5 - 3+a=a - 8\).

Step3: Set the left - hand and right - hand limits equal

Since the function is continuous at \(x = 1\), \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)\).
So \(11=a - 8\).
Solve for \(a\): \(a=11 + 8=19\).

Answer:

\(19\)