Question

The approximate value of $f(5.001)$, where $f\left(x\right)=x^3-7x^2+15$ is

• A. $-34.995$
• B. $-33.995$
• C. $-33.335$
• D. $-35.995$

Answer 1

The correct option is A

$-34.995$

Explanation for the correct option:

$Letx=5and∆x=0.001$

Then, we have

$\begin{array}{rcl}f(5.001)& =& f(x+∆x)\\ & =& {(x+∆x)}^3-7{(x+∆x)}^2+15\end{array}$

Now, $\begin{array}{rcl}∆y& =& f(x+∆x)-f\left(x\right)\end{array}$

$\begin{array}{rcl}\therefore f(x+∆x)& =& f\left(x\right)+∆y\\ & \approx & f\left(x\right)+f^{\text{'}}\left(x\right)∆x[\because dx=∆x]\end{array}$

$$\begin{gathered} \Rightarrow f(5.001)\approx (x^3-7x^2+15)+(3x^2-14x)∆x \\ \Rightarrow \left[{\left(5\right)}^3-7{\left(5\right)}^2+15\right]+\left[3{\left(5\right)}^2-14\left(5\right)\right](0.001)[\because x=5,∆x=0.001] \\ \Rightarrow (125-175+15)+(75-70)(0.001) \\ \Rightarrow -35+\left(5\right)(0.001) \\ \Rightarrow -34.995 \end{gathered}$$

Hence, the correct answer is option (A).