Question

Find the approximate value of $f(3.01)$, where $f(x)=3x^2+3$.

• (A) 30.18
• (B) 30.018
• (C) 30.28
• (D) 30.08

Answer 1

The Correct Option is A

Explanation:
Let, small charge in $x$ be $\mathrm{\Delta }x$ and the corresponding change in $y$ is $\mathrm{\Delta }y$.$\mathrm{\Delta }y=\frac{dy}{dx}\mathrm{\Delta }x=f^{\prime }(x)\mathrm{\Delta }x$Now that $\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)$Therefore, $f(x+\mathrm{\Delta }x)=f(x)+\mathrm{\Delta }y$Given: $f(x)=3x^2+3$Let, $x+\mathrm{\Delta }x=3.01=3+0.01$Therefore, $x=3$ and $\mathrm{\Delta }x=0.01$$f(x+\mathrm{\Delta }x)=f(x)+\mathrm{\Delta }y$$\Rightarrow f(x+\mathrm{\Delta }x)=f(x)+f^{\prime }(x)\mathrm{\Delta }x$$\Rightarrow f(3.01)=3x^2+3+(6x)\mathrm{\Delta }x$$\Rightarrow f(3.01)=3(3{)}^2+3+(6\cdot 3)(0.01)$$\Rightarrow f(3.01)=30+0.18$$\Rightarrow f(3.01)=30.18$Hence, the correct option is (A).