Find the values of a, b and c that make the function f continuous at x = 1, where f(x) = (a x^2 - x - 1)/(x - 1) for x > 1, f(1) = b, and f(x) = -x + c for x < 1.

Step-by-step solution

1. 1

   State what continuity at a point actually demands. Three separate quantities must exist and agree:

   $\lim_{x\to1^-}f(x)=f(1)=\lim_{x\to1^+}f(x)$

   Because the three branches carry three unknowns, this gives exactly the three equations needed. Work from the branch that constrains the most — the right-hand one.

2. 2

   Force the right-hand limit to be finite; this is what determines $a$. As $x\to1^+$ the denominator $x-1\to0$. If the numerator did not also tend to $0$, the quotient would blow up to $\pm\infty$ and no choice of $b$ or $c$ could rescue continuity. So we need

   $a(1)^2-1-1=0\quad\Longrightarrow\quad a-2=0\quad\Longrightarrow\quad a=2.$

   For example $a=3$ leaves numerator $\to1$ over denominator $\to0^+$, giving $+\infty$: no repair is possible.

3. 3

   Cancel the common factor and evaluate the limit. With $a=2$ the numerator factors because $x=1$ is a root:

   $2x^2-x-1=(2x+1)(x-1)$

   $\lim_{x\to1^+}\frac{(2x+1)(x-1)}{x-1}=\lim_{x\to1^+}(2x+1)=3$

   Cancelling is legitimate here because the limit only concerns $x\neq1$, where $x-1\neq0$.

4. 4

   Match the value at the point, which determines $b$. Continuity requires $f(1)$ to equal that limit:

   $b=3.$

5. 5

   Match the left-hand limit, which determines $c$. The left branch is a polynomial, so its limit is its value:

   $\lim_{x\to1^-}(-x+c)=-1+c$

   $-1+c=3\quad\Longrightarrow\quad c=4.$

6. 6

   Verify all three quantities agree. With $a=2$, $b=3$, $c=4$: the left limit is $-1+4=3$, the value is $f(1)=3$, and the right limit is $2(1)+1=3$. All equal, so $f$ is continuous at $x=1$. Numerically, $f(0.999)=3.001$ and $f(1.001)=3.002$, closing in on $3$ from both sides as expected.