Problem 18 - Calculus Single Variable

Question

Find a point \(x\) where \(f^{\prime}(x)=6\). $$ f(x)=3 x^{2}-6 x+50 $$

Step-by-Step Solution

Answer

The point where \( f^{'}(x) = 6 \) is \( x = 2 \).

Step 1: Find the Derivative

To find the point where the derivative equals 6, we need to first calculate the derivative of the function. The given function is \( f(x) = 3x^2 - 6x + 50 \). The derivative \( f^{ ext{'} }(x) \) is found by taking the derivative of each term: the derivative of \( 3x^2 \) is \( 6x \), the derivative of \( -6x \) is \( -6 \), and the derivative of a constant is 0. Thus: \[ f^{ ext{'}}(x) = 6x - 6 \]

Step 2: Set the Derivative Equal to 6

Now that we have the derivative \( f^{ ext{'}}(x) = 6x - 6 \), we set it equal to 6 to find the point where the derivative is 6: \[ 6x - 6 = 6 \]

Step 3: Solve the Equation

Solve the equation from Step 2 to find the value of \( x \). Add 6 to both sides to get:\[ 6x = 12 \].Now, divide both sides by 6:\[ x = 2 \].

Step 4: Confirm the Solution

We determined that \( x = 2 \) is the point where \( f^{ ext{'}}(x) = 6 \). To confirm, substitute \( x = 2 \) back into the derivative and check: \[ f^{ ext{'}}(2) = 6(2) - 6 = 12 - 6 = 6 \].The solution \( x = 2 \) satisfies the condition \( f^{ ext{'}}(x) = 6 \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing. It involves calculating the derivative. The derivative represents the slope of the tangent line to the function at any given point. To differentiate a function, you systematically apply rules that help translate the function's behavior into its rate of change.

For example, given the function \( f(x) = 3x^2 - 6x + 50 \), finding the derivative involves applying basic differentiation rules. Remember that:

• The derivative of \( ax^n \) is \( n \cdot ax^{n-1} \).
• The derivative of a constant is zero.
• Derivatives are linear, meaning you can differentiate terms individually and add the results.

Using these rules, the derivative of \( f(x) \) becomes \( f'(x) = 6x - 6 \). Thoroughly understanding differentiation is essential because it helps identify how function values move or change with respect to input values.

Critical Points

Critical points are key values in calculus that indicate where the derivative of a function is zero or undefined. These points can represent local maxima, minima, or saddle points. Finding critical points is crucial when analyzing functions, as they provide insights into the function's behavior and can help in optimization problems.

However, in the context of this exercise, we're looking for a specific value of the derivative, not just zero. We set \( f'(x) = 6 \) because the problem instructs us to locate where the derivative equals this particular value. Solving the equation \( 6x - 6 = 6 \) gives you the critical point \( x = 2 \). Recognize that critical points reveal much more:

• Points where the function changes direction.
• Potential turning points, signaling relative extremes of the function's graph.

Though our exercise didn't revolve around finding where the derivative is zero, knowing how to find critical points remains an indispensable calculus skill.

Quadratic Function

Understanding quadratic functions is straightforward. A quadratic function is a polynomial function of degree 2, typically in the form \( ax^2 + bx + c \). Graphically, it is represented as a parabola, which may open upwards or downwards depending on the leading coefficient \( a \).

The example function \( f(x) = 3x^2 - 6x + 50 \) is a quadratic function with:

• Leading coefficient \( a = 3 \), opening the parabola upwards.
• Constant term \( c = 50 \), determining the vertical position of the parabola.

An essential aspect of quadratic functions is locating their vertex or their axis of symmetry, often found via the formula \( x = -\frac{b}{2a} \). While this exercise did not require finding the vertex, knowing its computation helps when analyzing other parts of the graph.
In our specific task, the focus lay on how derivatives interact with the quadratic term to inform us about the curve's growth. By differentiating, we captured how changes in \( x \) affect \( f(x) \), allowing us to determine points of specific slopes, such as where the derivative equals 6. This process highlights the dynamic relationship between calculus and quadratic functions.