Question

The approximate value of $\sqrt[5]{{33}}$ correct to $4$ decimal places is
A) $2.0000$
B) $2.1001$
C) $2.0125$
D) $2.0500$

Answer 1

Hint: According to the mean value theorem, if the function is continuous on \[\left[ {a,b} \right]\] and differentiable on \[\left( {a,b} \right)\], then by the mean-value theorem we have: $f(b) - f(a) = f'(c)(b - a),c \in \left[ {a,b} \right]$ .
In the question, we need to determine the approximate value of $\sqrt[5]{{33}}$ upto 4 decimal places. Here at first, to compare the given - we form a function $y$ then, $f(y) = {x^{\left( h \right)}}$ and then, we can rewrite, as $x = 33$ as $f(y) = {\left( {32 + 1} \right)^{\left( {\dfrac{1}{5}} \right)}}$ and apply to Mean Value Theorem.

Complete answer:
Let us consider the $\sqrt[5]{{33}}$ as a function of $x$ then, \[f(x) = {(x + 1)^h}\] where, $x = 32$ and $h = \dfrac{1}{5}$ .
Now applying Mean Value Theorem, we have:
Also here, $a = x$ and $b = x + h$ for some \[h > 0\] (since $b > a$)
 We have\[c \in [a,b]\quad \Rightarrow \quad c = x + \theta h\qquad {\text{for some }}\theta \in (0,1).\]]
Using the property - \[f(a + b) = f(a) + hf'(a)\] we get,
\[{\left( {32 + 1} \right)^{\dfrac{1}{5}}} = {(32)^{\dfrac{1}{5}}} + \dfrac{1}{5} \times \left( {\dfrac{1}{{{{(32)}^{\dfrac{4}{5}}}}}} \right)\]
Simplify the above equation -
\[
  \sqrt[5]{{33}} = 2 + \dfrac{1}{{80}} \\
   = 2 + 0.0125 \\
   = 2.0125 \\
 \]
Therefore, the approximate value of $\sqrt[5]{{33}}$correct to$4$ decimal places is $2.0125$

Hence, option C is the correct answer.

Additional Information:In particular, if the continuous function is given on the closed interval [a, b] and the differentiable on the open interval (a, b), then there always exists the point c in (a, b) such a way that the tangent at C is parallel to the secant line through the endpoints (a, f(a)) and (b, f(b)) and that is – $f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$

Note: In such types of problems first we need to convert the given number as a function of \[y = f(x)\], then we use Roll’s Mean Value Theorem which states that if a function \[f\] is continuous on the closed interval \[\left[ {a,{\text{ }}b} \right]\]and differentiable on the open interval \[\left( {a,{\text{ }}b} \right)\]such that, \[f\left( a \right){\text{ }} = {\text{ }}f\left( b \right)\] then \[f\prime \left( x \right){\text{ }} = {\text{ }}0\] for some \[x\] with \[a{\text{ }} \leqslant {\text{ }}x{\text{ }} \leqslant {\text{ }}b\]. Then put the value in \[\therefore f(a + b) = f(a) + hf'(a)\]and find the final solution.