Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{f}(\mathrm{x})=[|\mathrm{x}|-\mid \mathrm{x}-1]]^{2}$
What is $\mathrm{f}^{\prime}(\mathrm{x})$ equal to when $0 < \mathrm{x} < 1 ?$
Options:
What is $\mathrm{f}^{\prime}(\mathrm{x})$ equal to when $0 < \mathrm{x} < 1 ?$
Solution:
1597 Upvotes
Verified Answer
The correct answer is:
$8 x-4$
Given $f(x)=(|x|-|x+1|)^{2}$
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
When $0 < x < 1$
$f(x)=(2 x-1)^{2}$
$f^{\prime}(x)=2(2 x-1) \cdot 2=4(2 x-1)$
$f^{\prime}(x)=8 x-4$
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
When $0 < x < 1$
$f(x)=(2 x-1)^{2}$
$f^{\prime}(x)=2(2 x-1) \cdot 2=4(2 x-1)$
$f^{\prime}(x)=8 x-4$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.