Problem 69 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the function \(f(x)=x^{5}-4 x+1\) has at least two zeros in the interval \((0,2)\).

Step-by-Step Solution

Answer

The function \(f(x) = x^5 - 4x + 1\) is continuous in the given interval (0,2). Calculating the function values at the interval bounds, we find that there is a sign change between f(0) and f(1) as well as between f(1.5) and f(2). Using the Intermediate Value Theorem, we conclude that there are at least two zeros in the interval (0,2).

Step 1: Check that the function is continuous in the given interval

In the given interval (0,2), the function \(f(x) = x^5 - 4x + 1\) is a polynomial function, and polynomial functions are continuous everywhere, including in the interval (0,2).

Step 2: Calculate function values at interval bounds

Find f(0) and f(2): \[f(0) = (0)^5 - 4(0) + 1 = 0 - 0 + 1 = 1\] \[f(2) = (2)^5 - 4(2) + 1 = 32 - 8 + 1 = 25\] We have f(0) = 1 and f(2) = 25.

Step 3: Check for sign changes between f(x) values

Here we see that there is no sign change between f(0) and f(2), so no zero is guaranteed between them. To find at least two zeros, let's now check for one zero somewhere between (0, 1) and another between (1, 2). Recall that f(0) > 0. Calculate f(1): \[f(1) = (1)^5 - 4(1) + 1 = 1 - 4 + 1 = -2\] Now, we have a sign change between f(0) and f(1). According to the Intermediate Value Theorem, there is at least one zero between (0,1).

Step 4: Check for another sign change

Calculate f(1.5): \[f(1.5) = (1.5)^5 - 4(1.5) + 1 ≈ -0.56\] There is no sign change between f(1) and f(1.5), so no zero is guaranteed between them. Therefore, we will calculate f(2) again for clarity: \[f(2) = (2)^5 - 4(2) + 1 = 32 - 8 + 1 = 25\] Now, we have a sign change between f(1.5) and f(2). According to the Intermediate Value Theorem, there is at least one zero between (1.5,2).

Step 5: Conclusion

The function \(f(x) = x^5 - 4x + 1\) shows a sign change in the intervals (0,1) and (1.5,2). Therefore, using the Intermediate Value Theorem, there are at least two zeros in the interval (0,2).

Key Concepts

Polynomial Function

A polynomial function is an expression involving variables raised to whole number powers, multiplied by coefficients. In this case, the polynomial function is given by:\[ f(x) = x^5 - 4x + 1 \]Understanding polynomial functions helps predict their general behavior, such as:

• They are smooth and continuous over their domain.
• Defined for all real numbers, making them easier to work with in different intervals.
• The degree of the polynomial (highest power of x) determines the 'shape' of the graph.

Here, the highest degree term is \(x^5\), shaping how the function behaves at positive and negative extremes. This characteristic is crucial when analyzing them in any given interval.

Continuity

Continuity of a function means there are no breaks, jumps, or holes in its graph over a particular interval. Polynomial functions, like \(f(x) = x^5 - 4x + 1\), are continuous everywhere, ensuring that they follow a smooth path along their curve.

• Because these functions are defined for every real number, they do not have breaks.
• In the interval (0,2), \(f(x)\) is continuous, ensuring applicability of analysis methods like the Intermediate Value Theorem.

The concept of continuity is essential to discovering certain features of polynomial functions, such as locating zeros.

Zeros of a Function

Zeros, or roots, of a function are the values of \(x\) at which the function equals zero. For example:\[ \text{If } f(c) = 0, \text{ then } c \text{ is a zero of } f(x). \]Finding zeros is often pivotal in algebra and calculus because they provide key insights into function behavior.

• In the exercise, we identify zeros to understand the intersection of the polynomial with the x-axis.
• The zeros in the interval (0,2) show places the function switches from positive to negative or vice versa.

Recognizing zeros allows for better understanding of where a function increases, decreases, and changes direction.

Sign Change

A sign change occurs when a function moves from positive to negative or vice versa. Observing sign changes is vital when working with continuous functions, such as polynomials, for locating zeros:

• In this context, we check the values of \(f(x)\) at various points in the interval to identify sign changes.
• In the exercise, \(f(x)\) showed a sign change between \(f(0)\) and \(f(1)\), indicating a zero exists between these points.
• Another sign change between \(f(1.5)\) and \(f(2)\) suggests a zero in \((1.5, 2)\).

Sign change analysis paired with continuity ensures a reliable method for predicting and confirming the existence of zeros, using the Intermediate Value Theorem.

Interval Analysis

Interval analysis involves evaluating a function within a specific range to understand its trends and specific points of interest, like zeros, maxima, or minima.

Importance of Interval Analysis

• Facilitates systematic examination of function behavior over specific domains.
• Helps determine features like whether the function crosses the x-axis (sign changes).
• Supports employing tools like the Intermediate Value Theorem to reach conclusions about zeros in polynomials.

Through interval analysis, the exercise demonstrated an effective approach to identifying two intervals, \((0,1)\) and \((1.5,2)\), where zeros exist, confirming the behavior of \(f(x)\) at those critical areas.