Question

If $\left \begin{array}{ll} f(x)= x^{2} + \alpha , & \quad for \ x \geq 0 \\ \hspace{0.75cm}= \sqrt[2]{x^{2}+1}+ \beta , & \quad for \ x \ \textless \ 0 \end{array} \right$ and f(1/2) = 2, is continuous at x=0, find α and β.

Answer 1

concept : any function, y = f(x) is continuous at x = a, only when

$\displaystyle\lim_{x\to a}f(x)=f(a)$

here, case 1 : x ≥ 0 , f(x) = x² + α

case 2 : x < 0, f(x) = √(x² + 1) + β

it is given that function is continuous at x = 0.

so, $\displaystyle\lim_{x\to 0}f(x)=f(0)$

or, $\displaystyle\lim_{x\to 0^+}=\displaystyle\lim_{x\to 0^-}=f(a)$

or,$\displaystyle\lim_{x\to 0^+}(x^2+\alpha)$ = $\displaystyle\lim_{x\to 0^-}\sqrt{x^2+1}+\beta=f(0)$......(1)

[ you did mistake , here f(0) = 2 not f(1/2) = 2]

now, $\displaystyle\lim_{x\to 0^+}(x^2+\alpha)=\alpha$....(2)

$\displaystyle\lim_{x\to 0^-}\sqrt{x^2+1}+\beta=1+\beta$.....(3)

from equations (1) ,(2) and (3),

α = 2 and 1 + β = 2 ⇒β = 1

Answer 2

Answer:

Step-by-step explanation: