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Chapter 2: Problem 78

Sketch the graph of the given function \(f\) on the interval [-1.3,1.3] $$ f(x)=-3 x^{3}+4 $$

Short Answer

Expert verified
To sketch the graph of the function \(f(x)=-3x^3+4\) on the interval \([-1.3, 1.3]\), first find the critical point at \(x=0\). The graph exhibits a hill-type curve around this critical point, based on the behavior of the function when \(x<0\) and \(x>0\). Plot the function values at the interval endpoints (\(x=-1.3\) and \(x=1.3\)). Now, draw a smooth curve connecting these points, keeping in mind the behavior near the critical points and the nature of the cubic function. The graph will start from the left endpoint, curve upward to the highest point at the critical point, and then curve back downward to the right endpoint.

Step by step solution

01

Find the critical points of the function.

To find the critical points, we first need to find the derivative of the given function. The derivative of a function will give us information about the slope of the function at different points, which in turn helps us determine the function's critical points. The derivative of the function \(f(x) = -3x^3 + 4\) can be found using basic power rules. The derivative of the function is: $$ f'(x) = \frac{d}{dx}(-3x^3 + 4) = -9x^2 $$ Now that we have the derivative, we can set it equal to zero to find the critical points: $$ -9x^2 = 0 $$
02

Solve for the critical points.

Now that we have set the derivative equal to zero, we can solve for the critical points. For the equation \(-9x^2 = 0\), we can solve for \(x\): $$ x^2 = 0 \Rightarrow x = 0 $$ We have one critical point at \(x = 0\).
03

Determine behavior around the critical point.

Let's find the behavior of the function around the critical point. When \(x < 0\), the function is positive (-3 times a negative number to the power of 3 will result in a positive result). When \(x > 0,\) the function is negative (-3 times a positive number to the power of 3 will result in a negative result). There is no undefined point in this function so we don't need to worry about its behavior at undefined points.
04

Identify the endpoints of the interval.

We have the critical point and the organized behavior around that point. Now let's focus on the endpoints given in the interval. The endpoints of our interval are \(x = -1.3\) and \(x = 1.3\). We can calculate the function values at these endpoints as follows. - For \(x = -1.3\): $$ f(-1.3) = -3(-1.3)^3 + 4 $$ - For \(x = 1.3\): $$ f(1.3) = -3(1.3)^3 + 4 $$ Evaluate these function values to plot them on the graph.
05

Sketch the graph.

Now that we have the critical point, behavior of the function around the critical point, and the endpoints of the interval, we can sketch the graph of the function on the interval \([-1.3, 1.3]\). 1. Plot the critical point at \(x = 0\). 2. Plot the function values at the interval endpoints (\(x = -1.3\) and \(x = 1.3\)). 3. Since the function is cubic, and given the behavior around the critical point, the graph will have a smooth transition and would be similar to a hill-type curve in the middle. Now you can draw the curve connecting these points, keeping in mind the behavior near the critical points and the nature of the cubic function. The graph will start from the left endpoint, curve upward with the highest point at the critical point, and then curve back downward to the right endpoint.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
In mathematics, critical points are essential in understanding the behavior of functions. They are the points where the derivative of the function is zero or undefined, indicating potential maxima, minima, or points of inflection. To identify critical points for a given function, you differentiate the function and set the derivative equal to zero. Solving this equation will give you the values of x that are the critical points.

For the function in the exercise, \( f(x) = -3x^3 + 4 \), the derivative is \( f'(x) = -9x^2 \). Setting this derivative equal to zero yields \( x^2 = 0 \) and subsequently \( x = 0 \), which is the sole critical point of this function. This is the point where the graph could have a peak or a trough鈥攃onstraints of the graph's behavior that are pivotal for its graphical representation.
Derivatives of Functions
The derivative of a function represents the rate at which the function's output value is changing at each point relative to its input value. In other terms, the derivative provides the slope of the tangent line to a curve at any given point. When we differentiate a constant, the result is zero because constants do not change, and therefore the slope is horizontal, or after differentiation, it becomes a zero.

In the context of our function, \( f(x) = -3x^3 + 4 \), taking the derivative gives \( f'(x) = -9x^2 \). This tells us how the function鈥檚 rate of change behaves depending on the value of \( x \). For all values of \( x \), except for the critical point, this derivative suggests a slope or a rate of change that isn't horizontal.
Cubic Functions
A cubic function is a polynomial of degree three. It has the general form \( ax^3 + bx^2 + cx + d \), where \( a \), \( b \), \( c \), and \( d \) are constants, and \( a \) is nonzero. These functions are known for their s-shaped curves and can possess up to two turns. They can either open upwards or downwards depending on the sign of the leading coefficient.

In our exercise, the function \( f(x) = -3x^3 + 4 \) is a cubic function with \( a = -3 \), \( b = c = 0 \) and \( d = 4 \). The negative leading coefficient indicates that the graph of this cubic function opens downwards. Thus, as \( x \) increases or decreases, the slopes of the tangent lines (given by the derivative) will decrease in magnitude because of the behavior of cubic functions.
Power Rule for Derivatives
The power rule is a basic principle used to differentiate functions that are monomials with the form \( cx^n \), where \( c \) is a constant, and \( n \) is a real number. According to the power rule, the derivative of such a function is \( ncx^{n-1} \). This simplicity is a powerful tool that simplifies the process of differentiation.

In the case of the function from our exercise, applying the power rule to \( -3x^3 \) gives us the derivative \( f'(x) = -9x^2 \). We derived this by multiplying the exponent 3 by the constant -3 and then subtracting 1 from the exponent. This rule is particularly handy because it allows for fast calculation of the derivatives without the need for more complex differentiation rules.

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Most popular questions from this chapter

Using the result that \(\sqrt{2}\) is irrational, explain why \(2^{1 / 6}\) is irrational.

Show that \(\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\)

Fermat's Last Theorem states that if \(n\) is an integer greater than \(2,\) then there do not exist positive integers \(x, y,\) and \(z\) such that $$ x^{n}+y^{n}=z^{n} $$ Fermat's Last Theorem was not proved until 1994, although mathematicians had been trying to find a proof for centuries. Use Fermat's Last Theorem to show that if \(n\) is an integer greater than \(2,\) then there do not exist positive rational numbers \(x, y,\) and \(z\) such that $$ x^{n}+y^{n}=z^{n} $$ [The equation \(3^{2}+4^{2}=5^{2}\) shows the necessity of the hypothesis that \(n>2\).]

Sketch the graph of the given function \(f\) on the interval [-1.3,1.3] $$ f(x)=3 x^{4} $$

Sketch the graph of the given function \(f\) on the interval [-1.3,1.3] $$ f(x)=x^{3}+1 $$
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