If fx= (x^2-16)/(x-4) if x>4 2x if x≤4 then f^(')4^(-)+f^(')4^(+)=

Correct answer: (C) 3. Given, fx= (x^2-16)/(x-4) if x>4 2x if x≤4 ⇒fx= x+4 if x>4 2x if x≤4 Now differentiating with respect to x we get, ⇒f^(')x= 1 if x>4 2 if x<4 So, f^(')4^(-)+f^(')4^(+)=2+1=3

Q107 / 160 · TS EAMCET 2022 (18 Jul Shift 1) · Mathematics

If $f (x) = \{\begin{matrix} \frac{x^{2} - 16}{x - 4} & if x > 4 \\ 2 x & if x \leq 4 \end{matrix}$ then $f^{'} (4^{-}) + f^{'} (4^{+}) =$

A
$1$
B
$2$
Correct answer
C
$3$
D
$4$

Solution

Given,
$f (x) = \{\begin{matrix} \frac{x^{2} - 16}{x - 4} & if x > 4 \\ 2 x & if x \leq 4 \end{matrix}$
$\Rightarrow f (x) = \{\begin{matrix} x + 4 & if x > 4 \\ 2 x & if x \leq 4 \end{matrix}$
Now differentiating with respect to $x$ we get,
$\Rightarrow f^{'} (x) = \{\begin{matrix} 1 & if x > 4 \\ 2 & if x < 4 \end{matrix}$
So, $f^{'} (4^{-}) + f^{'} (4^{+}) = 2 + 1 = 3$

Hence, the correct answer is option: 'C'.

If fx=x2-16x-4 if x>42x if x≤4 then f'4-+f'4+=

Solution

Relevant Questions

If $f (x) = \{\begin{matrix} \frac{x^{2} - 16}{x - 4} & if x > 4 \\ 2 x & if x \leq 4 \end{matrix}$ then $f^{'} (4^{-}) + f^{'} (4^{+}) =$