Problem 61 - Calculus Single Variable

Question

Let \(f\) be differentiable on, \(\mathbb{R}\). Suppose that \(f^{\prime}(2)>0\). Is \(f(2.000001)>f(2) ?\) Explain your answer.

Step-by-Step Solution

Answer

Yes, because \(f\) is increasing near 2 and \(f'(2) > 0\).

Step 1: Understanding the Derivative

The derivative of a function, denoted as \(f'(x)\), represents the rate of change or the slope of the function at a particular point \(x\). If \(f'(x) > 0\), it indicates that the function is increasing at that point.

Step 2: Analyzing the Given Information

We know that \(f'(2) > 0\). This implies that the function \(f\) is increasing at \(x = 2\). Therefore, for values of \(x\) slightly greater than 2, we expect \(f(x)\) to be greater than \(f(2)\).

Step 3: Considering Increment around 2

Given the continuity and differentiability of \(f\), for a very small increment like from \(2\) to \(2.000001\), since \(f'(2) > 0\), \(f(2.000001)\) should be slightly greater than \(f(2)\). The positive derivative means there's a small but steady increase.

Key Concepts

Rate of Change

The rate of change of a function at a particular point is an essential concept in understanding derivatives. Imagine driving along a road and checking your speedometer. The number it shows is similar to the rate of change for the car's position — your speed at that precise moment.

In calculus, this momentary speed at point \(x\) for a function \(f(x)\) is captured by its derivative \(f'(x)\). The derivative describes how fast the function's value is changing at that specific location. If \(f'(2) > 0\) for example, it indicates that the function is increasing — much like rising speeds when going downhill.

This positive rate suggests that as \(x\) gets slightly bigger, \(f(x)\) will too. Hence, around \(x = 2\), with \(f'(2) > 0\), we can expect a small growth in function values for a small increment in \(x\). This ties into the exercise: since \(f'(2)\) is positive, \(f(2.000001) > f(2)\) due to the increasing nature of the slope in that area.

Differentiability

Differentiability at a point means that the function has a well-defined tangent line at that point, and thus, a derivative. In simpler terms, the function behaves nicely — without sharp turns or breaks — allowing us to calculate its slope or rate of change at that precise place.

When we say a function \(f\) is differentiable on \(\mathbb{R}\), it means it has a derivative everywhere along the real number line. This ensures the function is smooth and continuous – key characteristics when analyzing behaviors like increasing or decreasing values.

Differentiability implies that small changes in the input \(x\) will produce proportionate and predictable changes in \(f(x)\). For the function \(f\) at \(x=2\), since \(f'(2) > 0\) and the function is differentiable, moving from \(2\) to \(2.000001\) guarantees \(f(x)\) will increase. This reaffirms why \(f(2.000001) > f(2)\).

Increasing Function

An increasing function, simply put, is a function that rises from left to right. The process by which a function climbs in value as the input \(x\) increases is primarily dictated by the sign of its derivative.

If \(f'(x) > 0\) for an interval, this suggests that for every small increase in \(x\), there is a corresponding increase in \(f(x)\). This property can be imagined like climbing up a gentle slope — each step you take forward causes you to rise slightly.

In our scenario, since \(f'(2) > 0\), the function is confirmed to be increasing around \(x=2\). This means that with even tiny trips forward along the x-axis, such as moving from \(2\) to \(2.000001\), the value of the function, \(f(x)\), will be higher. Consequently, this results in \(f(2.000001) > f(2)\), as expected from an increasing function.