Answer to Question #244081 in Calculus for Princewill Unuigbe

Question

1. (a) Give the definition of continuity of f at a. (b) Give the definition of the derivative of f at a.
2. For the function f(x) =  x 2 − x + 1 1 − x − x 2 if x < 0, if x ≥ 0

(a) Use your definition of continuity in 1(a) and the properties of limits to determine if f is continuous at a = 0. Show your work.

(b) Use your definition of derivative in 1(b) and the properties of limits to determine if f has a derivative at a = 0. Show your work.

Answer 1

1(a) A function f(x) is said to be continuous at a point x = a, in its domain if the following three conditions are satisfied: f(a) exists (i.e. the value of f(a) is finite) $\lim\limits_{x\rightarrow a} f(x)$ exists (i.e. the right-hand limit = left-hand limit, and both are finite) $\lim\limits_{x\rightarrow a} f(x)=f(a)$

1(b) For $y=f(x)$ , the derivative of  y  (with respect to  x  ) is

 

$\begin{aligned} &y^{\prime}=\frac{d y}{d x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \\ &f^{\prime}(x)=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \\ &f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &f^{\prime}(x)=\lim _{u \rightarrow x} \frac{f(u)-f(x)}{u-x} \end{aligned}$

2(a) Given that $f(x) = \begin{cases} x^2-x+1 &\text{if } x<0 \\ 1-x-x^2 &\text{if } x\geq0 \end{cases}$

Calculating left hand limit:

$\lim\limits_{x\rightarrow 0^-} f(x)\\ =\lim\limits_{x\rightarrow 0^-} (x^2-x+1)\\ =0-0+1\\ =1$

Calculating right hand limit:

$\lim\limits_{x\rightarrow 0^+} f(x)\\ =\lim\limits_{x\rightarrow 0^+} (1-x-x^2)\\ =1-0-0 \\ =1$

Since, LHL=RHL=f(0)

Therefore, f(x) is continuous at x = 0.

2(b) Differentiating f(x) w.r.t. (x), we get:

$f'(x) = \begin{cases} 2x-1 &\text{if } x<0 \\ -1-2x &\text{if } x\geq0 \end{cases}$

Calculating left-hand derivative, we get:

$\lim\limits_{x\rightarrow 0^-} f'(x)\\ =\lim\limits_{x\rightarrow 0^-} (2x-1)\\ =0-1\\ =-1$

Calculating right-hand derivative, we get:

$\lim\limits_{x\rightarrow 0^+} f'(x)\\ =\lim\limits_{x\rightarrow 0^+} (-1-2x)\\ =-1-0\\ =-1$

Since, LHD=RHD

Therefore, f(x) is differentiable at x = 0.