Problem 15 - Calculus Single Variable

Question

Find a point \(x\) where \(f^{\prime}(x)=6\). $$ f(x)=x^{2}+5 $$

Step-by-Step Solution

Answer

The point is at \( x = 3 \).

Step 1: Differentiate the function

To find where the derivative equals 6, first find the derivative of the function. Given the function \( f(x) = x^2 + 5 \), the derivative \( f'(x) \) is found using the power rule. The power rule states \( \frac{d}{dx}(x^n) = nx^{n-1} \). Applying it, we get \( f'(x) = 2x \), since the derivative of a constant is zero.

Step 2: Set derivative equal to 6

Now, set the derivative equal to 6 to find the x-value at which this occurs. This gives us the equation \( 2x = 6 \).

Step 3: Solve for x

Solve the equation from the previous step to find the value of \( x \). Divide both sides by 2, which gives \( x = 3 \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. Think of it as the slope of a tangent line to the curve of a function at a specific point. They help us understand how a function behaves and allows us to solve a variety of problems involving rates of change.

In our example with the function \( f(x) = x^2 + 5 \), the goal is to find the rate of change at any given point. By finding its derivative, we can determine how fast or slow the function's output is changing as input \( x \) changes. Calculating the derivative is often the first step to solving many calculus problems, especially when dealing with optimization or motion-related questions.

Applying the Power Rule

The power rule is a straightforward and extremely useful tool in calculus for differentiating functions of the form \( x^n \). It states: the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \).

Let's apply this to our function \( f(x) = x^2 + 5 \). Notice the term \( x^2 \). According to the power rule, differentiating \( x^2 \) with respect to \( x \) equals \( 2x^{2-1} = 2x \).

• A helpful hint: any constant added or subtracted in a function, like \(+5\) here, differentiates to zero, since constants don't change. Their rate of change is non-existent.

Thus, the derivative of \( f(x) = x^2 + 5 \) is simply \( 2x \). This simplifies our process, allowing us to focus on the non-constant terms to find how the function \( f \) behaves.

Finding Critical Points

Critical points in a function are the values of \( x \) where the derivative \( f'(x) \) is zero or undefined. These points are crucial as they often represent maxima, minima, or points of inflection where the function changes its behavior.

However, in this exercise, we're tasked with finding the point where \( f'(x) = 6 \), not zero. This is a bit different but helps illustrate how derivatives find specific behavior in functions. By setting our calculated derivative \( 2x \) equal to 6 and solving, we determine that at \( x = 3 \), the rate of change matches the desired value of 6.

• This isn’t a critical point in the traditional sense, but it does show where a specific rate of change occurs along the function.

Understanding how to find and interpret these points allows us to better grasp the nuances in a function's graph.