Application of Derivatives — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 6.4, Q.2,Page 216
Question

Find the approximate value of $f\left( {2.01} \right)$, where $f\left( x \right) = 4{x^2} + 5x + 2$.

Step-by-step Solution

We have,
$f\left( x \right) = 4{x^2} + 5x + 2 \Rightarrow f'\left( x \right) = 8x + 5$

Also, $f\left( {x + \Delta x} \right) \approx f\left( x \right) + \Delta xf'\left( x \right)$

Taking $x = 2$ and $\Delta x = 0.01$, we get
$f\left( {2.01} \right) \approx f\left( 2 \right) + \left( {0.01} \right)f'\left( 2 \right)$
$= \left( {4 \times {2^2} + 5 \times 2 + 2} \right) + \cfrac{1}{{100}}\left( {8 \times 2 + 5} \right) = 28.21$

$\Rightarrow f\left( {2.01} \right) \approx 28.21$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.