Increasing and Decreasing Functions

Solved Problems

Solution.

First we take the derivative:

Note that \(f^\prime\left( x \right) = 0\) at \(x = -2.\) The function and the derivative are undefined at \(x = 0,\) so there are two points that divide the real line (see figure above).

Using the interval method, we determine the sign of the derivative. We see that the derivative is positive everywhere on \(\left( { - \infty , - 2} \right]\) and \(\left( {0, +\infty} \right),\) and thus the function \(f\left( x \right)\) is increasing on \(\left( {-\infty,-2} \right] \cup \left( {0,+\infty} \right)\).

Similarly, we find that the function \(f\left( x \right)\) is decreasing on \(\left[ { - 2,0} \right).\)

Solution.

This function is defined and differentiable for all \(x \ge 0,\) except at the point \(x = 1\) where it has a discontinuity. Find the derivative \(f'\left( x \right)\) and determine the intervals where the derivative has a constant sign:

Using the interval method (Figure \(14\)), we see that the derivative is negative everywhere except at the points \(x = 0\) and \(x = 1\) where it does not exist.

Thus, the function is strictly decreasing in the intervals \(\left( {0,1} \right)\) and \(\left( {1,\infty} \right).\) Its schematic view is shown in Figure \(15.\)

Solution.

The function is defined and differentiable for \(x \gt 0.\) Its derivative is given by

Equate the derivative to zero and find the intervals where it has a constant sign:

The solution of the strict inequality \(f'\left( x \right) \gt 0\) is the infinite interval \(x \in \left( {{\frac{1}{e}},\infty } \right)\) (Figure \(16\)), and the solution of the inequality \(f'\left( x \right) \lt 0\) is the finite interval \(x \in \left( {0, {\frac{1}{e}}} \right).\)

Hence, by the sufficient criterion of monotonicity, the function is strictly increasing for \(x \in \left( {{\frac{1}{e}},\infty } \right)\) and strictly decreasing for \(x \in \left( {0, {\frac{1}{e}}} \right).\) Its view is schematically shown in Figure \(17.\)

Solution.

The function is defined and differentiable on the whole real line. Its derivative has the form:

The signs of the derivative can be easily determined by the interval method (Figure \(18\)). As it can be seen,

According to the sufficient criterion of monotonicity, the function is strictly increasing in the interval \(\left( {0,2} \right)\) and, accordingly, is strictly decreasing in the intervals \(\left( {-\infty,0} \right)\) and \(\left( {2, \infty} \right).\) Its schematic view is shown in Figure \(19.\)

Solution.

The function is defined on the following interval:

The derivative of this function is given by

Consequently,

Examine the sign of the derivative (Figure \(20\)).

Based on the sufficient condition for monotonicity, we conclude that the function is increasing for \(x \in \left( {0,{\frac{1}{2}}} \right)\) and decreasing for \(x \in \left( {{\frac{1}{2}},1} \right).\) The graph of the function is a half-circle centered at the point \(\left( {{\frac{1}{2}},0} \right)\) and radius \({\frac{1}{2}}\) (Figure \(21\)).

Solution.

The function is defined over all \(x\) except \(x = 0\) where it has a discontinuity.

Take the derivative:

Determine where the derivative is negative. Notice that the denominator is always positive if \(x \ne 0.\) Then we can write

So the function is strictly decreasing on \(\left( { - 1,0} \right)\) and \(\left( { 0,1} \right).\)

Solution.

This function is defined and differentiable for all \(x \in \mathbb{R},\) except for the values in which \(\cos x\) is equal to zero, that is except for \(x = {\frac{\pi }{2}} + \pi n,\) \(n \in \mathbb{Z}.\) Find the derivative \(f'\left( x \right):\)

Solve the inequality \(f'\left( x \right) \le 0:\)

Thus, the given function is decreasing on the whole set of real numbers, except at the points \(x = {\frac{\pi }{2}} + \pi n,\) \(n \in \mathbb{Z},\) where it is not defined.

Solution.

The function is defined for all \(x\) such that

Take the derivative using the quotient rule:

Locate the critical values of the function:

Determine the sign of the derivative in the open intervals we identified (see the sign chart above.)

The function is increasing if \(f^\prime\left( x \right) \ge 0\) and decreasing if \(f^\prime\left( x \right) \le 0.\) Therefore we can write the following answer:

Solution.

The function is defined for all \(x\) except \(x = 0.\)

The derivative of the function is given by

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

Using the interval method we find that

Consequently, we get the following answer:

Solution.

The function is defined and differentiable on the whole real line \(x \in \mathbb{R}.\) Its derivative has the following form:

We take a perfect square in this expression:

The function is strictly increasing when \(f'\left( x \right) \ge 0.\) Consequently, the condition of strict increasing for all \(x \in \mathbb{R}\) is written as

Obviously, the derivative takes the minimum value when \(x = 2.\) At this point it is equal:

From this we find the values of the parameter \(a,\) for which the function is strictly increasing:

Solution.

The cubic function is defined and differentiable on the whole real line. Its derivative is given by

Equating the derivative to zero, we determine the intervals of monotonicity of the function (Figure \(24\)):

Thus, at the transition (from left to right) through the point \(x = 1\), the function changes from increasing to decreasing, i.e. \(x = 1\) is the maximum point of the function. Similarly, \(x = 3\) is the minimum point of the function.

The cubic equation will have three distinct real roots in the case shown in Figure \(25.\) The maximum of the function must take a positive value, and the minimum must take a negative one. Thus, we arrive at the following condition:

Compute the values of the function \(f\left( x \right)\) at these points:

Now it is easy to find the range of values of the parameter \(a:\)

Solution.

The function in the left-hand side is defined and differentiable on the whole real line. The derivative of the function is given by

Calculate the roots of the derivative:

In the transition from left to right through the point \({x_1} = - 2\), the sign of the derivative changes from "\(+\)" to "\(-\)" (Figure \(26\)). Therefore the point\({x_1}\) is the maximum point. Similarly, we state that the point \({x_2} = 2\) is the minimum point.

Write down the values of the function at these points:

The equation has a single root if the points of maximum and minimum are on the same (upper or lower) half-plane, as shown in Figure \(27.\) Such configuration is described by the condition \({f_{\max }} \cdot {f_{\min }} \gt 0.\) Solving this inequality, we obtain:

Now we consider the case when the equation has \(2\) roots. A possible location of the graph of the function for this case is given in Figure \(28.\) Since one of the roots coincides with the point of maximum or minimum, then the corresponding condition is written as

Finally, the equation has three distinct roots, if the maximum and minimum points lie in the different half-planes (Figure \(29\)). Hence, here we have

Collecting together these results we can write the final answer:

The given cubic equation has one root when \(\left| a \right| \gt 16,\) two roots if \(\left| a \right| = 16\), and three roots when \(\left| a \right| \lt 16.\)