Q65 / 180 · COMEDK 2024 (Memory Based) · Mathematics
Find the approximate value of function $f(x)=x^3-5 x+10$ at $x=3.001$.
- A31.023
- B22.007
- C31.23
- Correct answerD22.07
Solution
We can write 3.001 as $x=3$ and $\Delta x=0.001$ and use the relation
$\begin{aligned}<br/>& f(x+\Delta x) \approx f(x)+\Delta x f^{\prime}(x) \\<br/>& f^{\prime}(x)=3 x^2-5<br/>\end{aligned}$
Now,
$\begin{aligned}<br/>f(3 & +0.001) \approx f(3)+(0.001)\left(f^{\prime}(5)\right) \\<br/>& \approx 27-15+10+0.001(75-5) \\<br/>& \approx 27-5+\frac{70}{1000} \\<br/>& \approx 22+0.07 \approx 22.07<br/>\end{aligned}$
$\begin{aligned}<br/>& f(x+\Delta x) \approx f(x)+\Delta x f^{\prime}(x) \\<br/>& f^{\prime}(x)=3 x^2-5<br/>\end{aligned}$
Now,
$\begin{aligned}<br/>f(3 & +0.001) \approx f(3)+(0.001)\left(f^{\prime}(5)\right) \\<br/>& \approx 27-15+10+0.001(75-5) \\<br/>& \approx 27-5+\frac{70}{1000} \\<br/>& \approx 22+0.07 \approx 22.07<br/>\end{aligned}$
Hence, the correct answer is option: 'D'.