Question

Graph carefully Graph the function \(f(x)=60 x^{5}-901 x^{3}+27 x\) in the window \([-4,4] \times[-10,000,10,000] .\) How many extreme values do you see? Locate all the extreme values by analyzing \(f^{\prime}\)

Short answer

In the given function \(f(x)=60x^5-901x^3+27x\), there are 3 extreme values: 1 local maximum and 2 local minimums. These extreme values are located at the critical points we found, which are approximately \(x_1\) = 0.2083, \(x_2\) = -0.2083, \(x_3\) = 1.533, and \(x_4\) = -1.533.

Step-by-step solution

Find the derivative of the function f(x)

To find the critical points, we need to find the first derivative of the function \(f(x)\). This will give us \(f'(x)\). Using the power rule for each term: \(f(x) = 60x^5 - 901x^3 + 27x\) \(f'(x) = 300x^4 - 2703x^2 + 27\)

Identify the critical points

Now, we need to find the critical points where \(f'(x)=0\). To find the values of \(x\) that make the derivative zero, we solve the equation: \(300x^4 - 2703x^2 + 27 = 0\) This is a quadratic equation in terms of \(x^2\). Make the substitution \(y = x^2\): \(300y^2 - 2703y + 27 = 0\) Now, we can use the quadratic formula to find the values of \(y\): \(y = \frac{-(-2703) \pm \sqrt{(-2703)^2-4(300)(27)}}{2(300)}\) Solve for y to get the two possible values: \(y_1 \approx 0.0434\) and \(y_2 \approx 2.3496\) Since we are interested in the values of \(x\), we must substitute back and solve for \(x\) using \(y = x^2\): For \(y_1\): \(x = \pm\sqrt{0.0434} \approx \pm 0.2083\) For \(y_2\): \(x = \pm\sqrt{2.3496} \approx \pm 1.533\) So, we have four critical points: \(x_1 \approx 0.2083\), \(x_2 \approx -0.2083\), \(x_3 \approx 1.533\), and \(x_4 \approx -1.533\)

Graph the function

Graph the function \(f(x)\) in the given window, \([-4, 4] \times [-10,000, 10,000]\). After graphing the function, observe the extreme values (local maximum and local minimum) and the turning points.

Identify the number of extreme values

Based on the graph of the function, we can see that there are 3 extreme values: 1 local maximum and 2 local minimums. The locations of the extreme values are at the critical points (\(x_1\), \(x_2\), \(x_3\), and \(x_4\)) we found in Step 2.

Key concepts

Finding the Derivative

Understanding how to find the derivative of a functioning calculus is crucial for analyzing its behavior. The process involves applying differentiation rules to determine rate of change or slope at any given point.
For example, when given a function like \(f(x) = 60x^5 - 901x^3 + 27x\), the derivative \(f'(x)\) represents the instantaneous rate of change of \(f(x)\) with respect to \(x\).

Using the Power Rule

To find the derivative, you can use the power rule, which simplifies the differentiation of terms like \(ax^n\). The derivative of \(ax^n\) is \(n \cdot ax^{n-1}\). So, for our function \(f(x)\), we differentiate each term separately to find the derivative, resulting in \(f'(x) = 300x^4 - 2703x^2 + 27\).

Power Rule Differentiation

The power rule is a basic differentiation technique in calculus and is used whenever you have a power of \(x\).
For any function of the form \(f(x) = ax^n\), the power rule states that the derivative is \(f'(x) = n \cdot ax^{n-1}\). It's a quick and effective method for finding derivatives of polynomial functions.

Applying the Rule

Applying the power rule to each term of the polynomial, we multiply the exponent with the coefficient and reduce the exponent by one. Employing this technique simplifies complicated polynomial functions. With practice, you will be able to calculate derivatives of functions like \(f(x) = 60x^5 - 901x^3 + 27x\) more efficiently, as seen in our example.

Solving Quadratic Equations

Quadratic equations take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are known values, and \(a\) is not equal to zero. The solutions to these equations are the 'roots' or 'x-intercepts' of the function.
To solve them, you can factor the equation, complete the square, or use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Application in Calculus

In calculus, when finding the derivative yields a quadratic equation, it's usually to determine the critical points of the original function. This step is essential as it helps in understanding the function's turning points which indicate maximums and minimums. In our problem, the substitution \(y = x^2\) converts the quartic equation into a quadratic, which we then solve using the quadratic formula.

Graphing Polynomial Functions

Graphing polynomial functions is a way to visualize the behavior of the equations over a range of values. These graphs reveal important features of the function, including intercepts, extreme values, and end behavior.
To graph a polynomial function, plot enough points to understand the shape, or use software or a graphing calculator, which is especially helpful for more complex polynomials like \(f(x) = 60x^5 - 901x^3 + 27x\).

Observing Critical Points

After plotting the function, you can observe its critical points, where the function either peaks or troughs, representing the local maximums and minimums. Our function, when graphed over the window \([-4,4] \times [-10,000,10,000]\), reveals three extreme values. These points correspond to the critical points found through solving for the derivative's roots.