Differentiation Question 126
Question: If f be a function given by $ f(x)=2x^{2}+3x-5. $ Then, $ f’(0)=mf’(-1), $ where m is equal to
Options:
A) -1
B) -2
C) -3
D) -4
Show Answer
Answer:
Correct Answer: C
Solution:
[c] We first find the derivatives of $ f( x ) $ at $ x=-1 $ and at $ x=0. $ We have, $ f’(-1)=\underset{h\to 0}{\mathop{\lim }}\frac{f(-1+h)-f(-1)}{h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{[2{{(-1+h)}^{2}}+3(-1+h)-5]-[2{{(-1)}^{2}}+3(-1)-5]}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{2h^{2}-h}{h}=\underset{h\to 0}{\mathop{\lim }}(2h-1)=2(0)-1=-1 $ and $ f’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{[2{{(0+h)}^{2}}+3(0+h)-5]-[2{{(0)}^{2}}+3(0)-5]}{h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{2h^{2}+3h}{h}=\underset{h\to 0}{\mathop{\lim }}(2h+3)=2(0)+3=3 $ Clearly, $ f’(0)=-3f’(-1) $
BETA
