Increasing and Decreasing Functions

Solved Problems

Solution.

A function is strictly increasing when the \(y-\)value increases as the \(x-\)value increases. One can see that the given function is strictly increasing on the intervals \(\left( { - 5, - 1} \right)\) and \(\left( {3,6} \right).\)

Solution.

According to the definition, a function is decreasing on an interval if \(f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)\) for any two points \({x_1} \le {x_2}.\)

Thus, a decreasing interval may also contain points where the function has a constant value. (This is not true for a strictly decreasing function.)

In our case, we see that the function is decreasing on the interval \(\left( { - 2,6} \right).\)

Solution.

We take two arbitrary points \({x_1}\) and \({x_2}\) such that

Consider the difference between the values of the function at these points:

It is obvious that in the last expression \({{x_2} - {x_1}} \gt 0\) and \({{x_2} + {x_1}} \gt 0\) (since, by assumption, only non-negative values of \(x\) are considered). As a result, we have

This means by definition that the function \(f\left( x \right) = {x^2} + 1\) is strictly increasing on the given interval.

Solution.

We choose two arbitrary points \({{x_1}}\) and \({{x_2}}\) such that \({x_1} \lt {x_2}.\) Consider the difference:

Factoring it as the difference of cubes, we obtain:

In the second bracket we can get a perfect square:

Hence it is clear that the quadratic expression is always positive (it is equal to zero only if \({x_1} = {x_2} = 0,\) which contradicts the condition \({x_1} \lt {x_2}.\))

Thus, \(f\left( {{x_2}} \right) - f\left( {{x_1}} \right) \gt 0,\) if \({x_2} - {x_1} \gt 0,\) i.e. the function \(f\left( x \right) = {x^3}\) is strictly increasing.

Solution.

This function is the sum of the functions \({x^4}\) and \(3{x^2}.\)

The first function \({x^4}\) can be considered as the product of two identical functions \({x^2}\). Example \(1\) shows that the quadratic function \({x^2}\) is strictly increasing for \(x \ge 0.\) Hence, the function \({x^4}\) is also strictly increasing for \(x \ge 0\) by the property \(4.\)

The second term \(3{x^2}\) is the triple sum of the functions \({x^2}\), so this term is also strictly increasing according to the property \(1.\)

Hence, the original function \(f\left( x \right) = {x^4} + 3{x^2}\) is the sum of two strictly increasing functions and consequently is also strictly increasing when \(x \ge 0.\)

Solution.

Let the points \({x_1}\), \({x_2}\) lie in the given interval \(\left[ {0,\pi } \right]\) and the following condition is satisfied: \({x_1} \lt {x_2}.\) Consider the difference:

Convert it by the cosine difference identity:

Since \({x_1},{x_2} \in \left[ {0,\pi } \right],\) then the half sum is bounded by the double inequality

Similarly, the half difference (assuming that \({x_2} \lt {x_1}\)) satisfies the inequality

For these ranges of angles, the sine is always positive. Therefore

Thus, the following relationship holds:

that is the cosine function is strictly decreasing on the interval \(\left[ {0,\pi } \right].\)

Solution.

Find the derivative:

Notice that the discriminant of the quadratic function is negative:

Therefore, the quadratic function has no zeros and has the same sign over the interval \(\left( { - \infty ,\infty } \right).\)

We choose \(x = 0\) to evaluate the sign of the derivative:

Hence, the function is strictly increasing on \(\mathbb{R}.\)

Solution.

Calculate the derivative:

The derivative is zero at the points \({x_1} = - 1,\) \({x_1} = 0,\) \({x_3} = 1.\)

Using the interval method we find the intervals where the derivative has a constant sign (see the sign chart above).

Hence, the function is increasing on \(\left( { - 1,0} \right)\) and \(\left( {1, + \infty } \right).\)

Solution.

We take the derivative and find the critical points:

Substituting the test value \(x = 0,\) we see that the derivative is negative for \(x \in \left( { - 1,5} \right).\) Hence, the function is decreasing on \(\left[ { - 1,5} \right].\) The length of the interval is \(L = 6.\)

Solution.

Find the derivative using the product rule:

Determine the sign of the derivative by the interval method. We see in the figure above that the derivative is positive for \(x \in \left( {0,4} \right),\) so the length of the interval on which the function is increasing is \(L = 4.\)

Solution.

The derivative of this function is given by

Determine the intervals where the derivative is positive and negative. Solve the following inequality:

Using the interval method we find that

Consequently, the function \(f\left( x \right) = {x^3} - 12x + 5\) is increasing (in the strict sense) in the intervals \(\left( { - \infty , - 2} \right)\) and \(\left( {2,\infty } \right)\) and, accordingly, is strictly decreasing in the interval \(\left( { - 2,2} \right).\)

Solution.

This function is defined and differentiable on the real line. Consider the inequality \(f'\left( x \right) \gt 0:\)

This inequality holds for all \(x\) except at the points where \(\cos x = -1,\) that is

However, if we consider the non-strict inequality \(f'\left( x \right) \ge 0,\) we obtain:

Thus, the function \(f\left( x \right) = x + \sin x\) is increasing (but not in the strict sense, i.e. the function is non-decreasing) for any \(x \in \mathbb{R}.\)

To check the result, we also consider the inequality \(f'\left( x \right) \lt 0:\)

This inequality has no solutions.

Solution.

The function is defined and differentiable on the whole set of real numbers. Calculate its derivative:

Determine the intervals where the derivative has a constant sign. Equate the derivative to zero and find the roots of the equation:

Determine the intervals where the derivative has a constant sign. Equate the derivative to zero and find the roots of the equation:

Thus, the function is decreasing (in the strict sense) in the intervals \(\left( { - \infty , - 1} \right)\) and \(\left( {1, \infty} \right)\) and increasing in the interval \(\left( {-1, 1} \right).\) Given that the root of the function is of the form \(x = 0,\) we can schematically draw its graph (Figure \(12\)).

Solution.

First we take the derivative of the function using the logarithmic differentiation:

The function \({x^x}\) is positive at \(x \gt 0,\) so the sign of the derivative is determined by the term \({\ln x + 1}.\)

Calculate the critical point:

Hence, the function is decreasing on \(\left( {0,\frac{1}{e}} \right]\) and increasing on \(\left[ {\frac{1}{e}, + \infty } \right).\)