Question

A function with the parameters \(a\) and \(b\) are given. Describe the critical points and possible points of inflection of \(f\) in terms of \(a\) and \(b\). \(f(x)=a x^{2}+b x+1\)

Short answer

Critical points are at \(x = -\frac{b}{2a}\); there are no points of inflection.

Step-by-step solution

Find First Derivative to Determine Critical Points

To find the critical points of the function, we first need the first derivative. Given the function \(f(x) = ax^2 + bx + 1\), the first derivative \(f'(x)\) is obtained by applying power rules. This yields \(f'(x) = 2ax + b\). Critical points occur where this derivative is zero or undefined. Since \(f'(x)\) is a linear polynomial, it is defined for all \(x\), so we set \(f'(x) = 0\) to find the critical points: \[2ax + b = 0\]

Solve for Critical Points

Solve the equation \(2ax + b = 0\) for \(x\). Rearranging gives: \[x = -\frac{b}{2a}\]These are the critical points expressed in terms of \(a\) and \(b\).

Find Second Derivative to Determine Concavity

To find possible points of inflection, determine the concavity by calculating the second derivative. Taking the derivative of \(f'(x) = 2ax + b\), the second derivative \(f''(x)\) is constant:\[f''(x) = 2a\]Since \(f''(x) = 2a\), it is independent of \(x\), meaning the concavity is constant.

Analyze Second Derivative for Inflection Points

Points of inflection occur where the sign of \(f''(x)\) changes. In this case, \(f''(x) = 2a\) is constant. If \(a > 0\), the function is concave up everywhere, and if \(a < 0\), it is concave down everywhere. There are no changes in concavity, thus no points of inflection.

Key concepts

First Derivative

The first derivative of a function gives us important insights into the function's behavior, specifically regarding critical points. With the given function, \(f(x) = ax^2 + bx + 1\), the first derivative, \(f'(x)\), is calculated using the power rule. This results in \(f'(x) = 2ax + b\).

• The critical points of a function are where the first derivative equals zero or is undefined.
• In our case, because \(f'(x)\) is a linear polynomial, it is defined for all \(x\).

Thus, to find the critical points, set \(f'(x) = 0\), which leads to the equation \(2ax + b = 0\). Solving for \(x\) gives us the critical point \(x = -\frac{b}{2a}\). This is the value of \(x\) where the slope of the tangent to the curve is zero, indicating a potential local maxima or minima in terms of \(a\) and \(b\). Also, remember:

• Critical points are candidates for local extrema or are where the function’s derivative changes its sign.
• Knowing \(a\) and \(b\) allows us to determine the precise location of these critical points.

Second Derivative

The second derivative provides critical information about the concavity of a function. For our function \(f(x) = ax^2 + bx + 1\), the first derivative is \(f'(x) = 2ax + b\). By taking the derivative of \(f'(x)\), we find the second derivative: \(f''(x) = 2a\).

• The second derivative test helps determine whether a critical point is a local maximum, a local minimum, or neither.
• Since our second derivative, \(2a\), is constant and not dependent on \(x\), it only depends on the value of \(a\).

This tells us:- If \(a > 0\), the second derivative is positive, meaning the function is concave up; any critical point is a local minimum.- If \(a < 0\), the second derivative is negative, indicating the function is concave down; any critical point is a local maximum.Therefore, \(f''(x)\) essentially confirms the nature of the critical points and the general shape of the parabola defined by the quadratic function. Understanding \(f''(x)\) is crucial in sketching the graph and predicting the behavior of \(f(x)\).

Inflection Points

Inflection points are where a function changes its concavity, which occurs where the second derivative changes its sign. For our function, the second derivative is \(f''(x) = 2a\).

• A change in the concavity of a function suggests that \(f''(x)\) should switch from positive to negative, or vice versa, at some point.
• This switch indicates the potential for an inflection point.

However, because \(f''(x) = 2a\) is constant:- The function maintains the same concavity across all its domain.- If \(a > 0\), the function is always concave up.- If \(a < 0\), it remains concave down.Given this, no switch occurs in \(f''(x)\), and therefore, there are no inflection points for this specific quadratic function. Recognizing this characteristic allows one to understand how this simple parabola behaves across its entire range.