Problem 10
Question
Find \(c\) so that \(f^{\prime}(c)=0 .\) \(f(x)=-x^{2}+4\)
Step-by-Step Solution
Verified Answer
The value of \(c\) is 0.
1Step 1: Differentiate the function
To find where the derivative is zero, we first need to find the derivative of the function. Given the function \( f(x) = -x^2 + 4 \), apply the power rule to differentiate. The derivative is \( f'(x) = -2x \).
2Step 2: Set the derivative equal to zero
Now that we have the derivative \( f'(x) = -2x \), set this expression equal to zero to find where the slope of the tangent line is zero: \(-2x = 0\).
3Step 3: Solve the equation for \(x\)
Solve the equation \(-2x = 0\) to find the value of \(x\). By dividing both sides by \(-2\), we have \( x = 0 \).
4Step 4: Interpret the result
The solution \( x = 0 \) indicates the input where the function’s derivative is zero. Thus, \( c = 0 \). This value of \( c \) is the critical point where the slope of the function's tangent is horizontal.
Key Concepts
Understanding DerivativesIdentifying Critical PointsApplying the Power Rule
Understanding Derivatives
The derivative of a function is a fundamental concept in calculus. It represents the rate of change of a function's output with respect to its input. Imagine you have a hill, and you want to know how steep it is at different points. Similarly, the derivative gives us a "slope" of the curve represented by the function.
In mathematical terms, if you have a function \( f(x) \), its derivative \( f'(x) \) tells you the slope of the tangent line at any given point on the function. This helps us understand how the function is behaving at that particular point.
In mathematical terms, if you have a function \( f(x) \), its derivative \( f'(x) \) tells you the slope of the tangent line at any given point on the function. This helps us understand how the function is behaving at that particular point.
- A positive derivative means the function is increasing.
- A negative derivative means the function is decreasing.
- A zero derivative indicates a constant slope, suggesting a possible peak, trough, or point of inflection, depending on the function.
Identifying Critical Points
A critical point of a function occurs where its derivative is zero or undefined. In simpler terms, these are points where the function's graph stops increasing or decreasing, or doesn't have a well-defined slope.
Those points are crucial because they often indicate points of maximum or minimum values, or subtle changes in the curve's shape. For example, consider \( f(x) = -x^2 + 4 \). When you take the derivative, \( f'(x) = -2x \), setting it equal to zero gives \( x = 0 \). This signifies a critical point.
By evaluating the second derivative or using other tests, you can determine the nature of the critical point:
Those points are crucial because they often indicate points of maximum or minimum values, or subtle changes in the curve's shape. For example, consider \( f(x) = -x^2 + 4 \). When you take the derivative, \( f'(x) = -2x \), setting it equal to zero gives \( x = 0 \). This signifies a critical point.
By evaluating the second derivative or using other tests, you can determine the nature of the critical point:
- If the function's derivative changes from positive to negative, you have a local maximum.
- If it changes from negative to positive, you have a local minimum.
- If it doesn’t change signs, the point might be neither a maximum nor minimum (usually indicating a saddle point or point of inflection).
Applying the Power Rule
The power rule is a quick and effective method for finding the derivative of a function that involves exponents. It states that if you have a function of the form \( x^n \), then its derivative is \( nx^{n-1} \).
Let's apply this to the function given in the exercise, \( f(x) = -x^2 + 4 \). Using the power rule:
Let's apply this to the function given in the exercise, \( f(x) = -x^2 + 4 \). Using the power rule:
- The term \( -x^2 \) becomes \( -2x^{2-1} = -2x \).
- The constant 4 becomes 0 because the derivative of a constant is always zero.
Other exercises in this chapter
Problem 9
In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=\frac{1}{\left(x^{3}-1\right)^{4}} $$
View solution Problem 10
Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a
View solution Problem 10
Find the derivative with respect to the independent variable. $$ f(x)=-3 \cos (1-2 x) $$
View solution Problem 10
Differentiate the functions with respect to the independent variable. \(f(x)=2 x e^{-3 x}\)
View solution