ii. suppose that $f(x)=\begin{cases}c - 2x^{2}&x<1\\cos(x - 1)&xgeq1end{cases}$, find number $c$ such that…

Explanation:

Step1: Recall continuity condition

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

Step2: Calculate left - hand limit

$\lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{-}}(c - 2x^{2})=c-2(1)^{2}=c - 2$.

Step3: Calculate right - hand limit

$\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1^{+}}\cos(x - 1)=\cos(1 - 1)=\cos(0)=1$.

Step4: Equate left - hand and right - hand limits

Since $f(x)$ is continuous at $x = 1$, we set $c-2=1$.

Step5: Solve for c

Adding 2 to both sides of the equation $c-2 = 1$, we get $c=1 + 2=3$.

Answer:

$c = 3$