How do you find a one-decimal place approximation for $-\sqrt[3]{35}$ ?

Hint: Here in this question we have been asked to find the approximate one-decimal place value for $-\sqrt[3]{35}$ . For answering this question we will assume a function $f\left( x \right)={{\left( 3+x \right)}^{3}}$ and then simplify it.

Complete step by step solution:
Now considering from the question we have been asked to find the approximate one-decimal place value for $-\sqrt[3]{35}$ .
For answering this question we will assume a function $f\left( x \right)={{\left( 3+x \right)}^{3}}$ and then simplify it.
Let us express $-{{\left( 35 \right)}^{\dfrac{1}{3}}}$ as a function
 $\begin{align}
  & f\left( x \right)={{\left( 3+x \right)}^{3}}\Rightarrow {{3}^{3}}+{{x}^{3}}+3{{\left( 3 \right)}^{2}}x+3\left( 3 \right){{x}^{2}} \\
\end{align}$
Let us assume that $x$ is so small, now we can say that $f\left( x \right)=27+27x$ .
Now let us consider that
 $\begin{align}
  & 35=f\left( x \right) \\
 & \Rightarrow 35={{\left( 3+x \right)}^{3}} \\
 & \Rightarrow {{\left( 35 \right)}^{\dfrac{1}{3}}}=\left( 3+x \right) \\
\end{align}$
So we need to find the value of $x$ then we will have
$\begin{align}
  & 35=27+27x \\
 & \Rightarrow 8=27x \\
 & \Rightarrow x=\dfrac{8}{27} \\
 & \Rightarrow x=0.3 \\
\end{align}$
Hence ${{\left( 35 \right)}^{\dfrac{1}{3}}}=3+0.3\Rightarrow 3.3$

Therefore we can conclude that the one-decimal approximation for $-\sqrt[3]{35}$ will be given as $-3.3$ .

Note: While answering questions of this type we should be sure with our concepts that we are going to apply in the process and the calculations that we are going to perform in between the steps. This is a very simple and easy question and can be answered accurately in a short span of time. Very few mistakes are possible in questions of this type. Here we should carefully select the function we are going to use because our answer depends on the function. So practice more questions for knowing the suitable questions. Here it is enough to write $\dfrac{8}{27}$ as 0.3 instead of 0.296 because we need only one-decimal approximation.