Question

if \({\bf{f'}}\left( {\bf{x}} \right){\bf{ = g'}}\left( {\bf{x}} \right)\) for\({\bf{0 < x < 1}}\), then \({\bf{f}}\left( {\bf{x}} \right){\bf{ = g}}\left( {\bf{x}} \right)\) for\({\bf{0 < x < 1}}\).

Short answer

The given statement is False

Step-by-step solution

Step:-1 Explanation

Consider the function \(f'\left( x \right) = g'\left( x \right)\) for\(0 < n < 1\).

Then \(f\left( x \right) = g\left( x \right)\) for \(0 < n < 1\).

Step:-2: Write the required concept

Let\({\bf{f:}}\left( {{\bf{0,1}}} \right) \to {\bf{R}}\)

\(f\left( x \right) = x + 6\)

And \({\bf{g:}}\left( {{\bf{0,1}}} \right) \to {\bf{R}}\)

\(g\left( x \right) = x + 1\)

Both f and g are polynomial functions.

\(\therefore \) it is continuous and differential.

Step:-3 Explanation for the question:

Now here we solve our question,

\(f'\left( x \right) = \frac{d}{{dx}}\left( {f\left( x \right)} \right) = \frac{d}{{dx}}\left( {x + 6} \right) = 1\)

\(g'\left( x \right) = \frac{d}{{dx}}\left( {g\left( x \right)} \right) = \frac{d}{{dx}}\left( {x + 1} \right) = 1\)

\( \Rightarrow f'\left( x \right) = g'\left( x \right)\) \(\forall x \in \left( {0,1} \right)\) or for \(0 < n < 1\).

But \(f\left( x \right) \ne g\left( x \right)\) for \(0 < n < 1\).

Because \(0 < x = \frac{1}{2} < 1\)

\(f\left( {\frac{1}{2}} \right) = \frac{1}{2} + 6 = \frac{{13}}{2}\)

\(g\left( {\frac{1}{2}} \right) = \frac{1}{2} + 1 = \frac{3}{2}\)

\(f\left( {\frac{1}{2}} \right) \ne g\left( {\frac{1}{2}} \right)\) at \(x = \frac{1}{2}\)

\(\therefore f\left( x \right) \ne g\left( x \right)\)for \(0 < n < 1\).

Hence our statement is false.