Question

Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.

5) If \(f'\left( x \right) < 0\) for \(1 < x < 6\), then f is decreasing on\(\left( {1,6} \right)\).

Short answer

The statement is true.

Step-by-step solution

Explanation

True given\(f'\left( x \right) < 0\)……….(1) on \(1 < x < 6\).

\(f\left( x \right)\)is differentiable.

\(\therefore \)it is continuous also on that interval.

Let \({x_1},{x_2} \in \left( {1,6} \right)\)

Such that\({x_1} > {x_2}\),

\( = {x_2} - {x_1} > 0\)

Apply the Mean value theorem

\(\therefore \)By mean value theorem

\(\frac{{f\left( {{x_1}} \right) - f\left( {{x_2}} \right)}}{{{x_2} - {x_1}}} = f'\left( c \right)\)

For some \(c \in \left( {1,6} \right)\)

From equation (1) \(f'\left( c \right) < 0\),

\(\frac{{f\left( {{x_1}} \right) - f\left( {{x_2}} \right)}}{{{x_2} - {x_1}}} < 0\)

\(f\left( {{x_2}} \right) < f\left( {{x_1}} \right)\)

Thus, we get if \({x_1} < {x_2}\)then:

\(f\left( {{x_2}} \right) < f\left( {{x_1}} \right)\)

\( = f\left( {{x_1}} \right) > f\left( {{x_2}} \right)\)

Which shows that \(f\left( x \right)\) is decreasing.

Hence, the question statement is true.