Let’s dive into the second derivative test, which is a super useful tool in calculus. It helps you figure out key points in a function, like where the graph changes direction or where it hits a peak or valley (also known as maxima and minima). If you're looking at a function with an independent variable (let’s say "x") and a dependent variable (we’ll call it "y"), the derivative helps you understand how the value of y changes as x changes. This is important for figuring out how a graph behaves, and you’ll definitely see it on tests like the SAT, ACT, or AP exams.

So, how does the second derivative test help? It tells you whether a point is a local maximum, minimum, or an inflection point (where the graph changes direction). If you get a positive second derivative at a point, it’s a minimum, and if it’s negative, it’s a maximum. This is super helpful for graphing functions and solving problems efficiently on exams like the SAT, ACT, and even the MCAT. So, mastering derivatives can really give you an edge when it comes to calculus questions!

Second Derivative Test

The second derivative test is a mathematical technique used to determine the nature of critical points and inflection points of a function. It involves examining the concavity of the function by evaluating its second derivative at the critical points.

If the second derivative is positive at a critical point, the function has a local minimum at that point. Conversely, if the second derivative is negative, the function has a local maximum. In addition, if the second derivative changes sign at a point, that point is an inflection point where the concavity of the function changes.

For a function say y=f(x);

The first derivative is calculated using the formula; \(\frac{dy}{dx}=f^{\prime}\left(x\right)\)

Similarly, the second derivative is calculated using the formula;\(\frac{d}{dx}f^{\prime}\left(x\right)=f^{^{\prime\prime}}\left(x\right)\).

Second Derivative Test For Maxima And Minima

Maxima and minima are understood as the extrema of a given function. Maxima and minima in general are the maximum or the minimum value of a function within the assigned set of ranges.

The second derivative test further depends on the sign of the second derivative at a given point. If the derivative is positive, the point denotes a minimum, and if the derivative is negative, the point denotes a maximum.

Mathematically saying;

If \(f′(c)=\frac{d}{dx}f(c)=0\) and\(f′′(c)=\frac{d^2}{dx^2}f(c)>0\) , then there is a local minimum at the point x=c.

If \(f′(c)=\frac{d}{dx}f(c)=0\) and \(f′′(c)=\frac{d^2}{dx^2}f(c)<0\), then there is a local maximum at the point x=c.

Steps To Find Local Extrema With Second Derivative Test

Now that we know the second derivative test for concave and convex function along with the formula. Let us understand the detailed steps on how to find local extrema with the second derivative test.

Step 1: To start with determining the first derivative f'(x) or \(\frac{d}{dx}f(x)\) of the given function i.e. f(x).

Step 2: Next, equate the obtained first derivative to zero i.e. f'(x) = 0 or \(\frac{d}{dx}f(x)=0\) and obtain the points(consider \(c_1\) and \(c_2\)).

Step 3: Now, determine the second derivative of the given function i.e. f”(x) or \(\frac{d^2}{dx^2}f(x)\).

Step 4: Lastly, substitute the points in the second derivative and estimate the local extrema. The condition for the same is;

If the value of the second derivative is greater than zero i.e.\(f′′(c_1)>0\ \text{ or }\frac{d^2}{dx^2}f(c_1)>0\), then the point (\(c_1\)) is said to be the local minima.

Similarly, if the second derivative is smaller than zero i.e.\(f′′(c_2)<0\text{ or }\frac{d^2}{dx^2}f(c_2)<0\), then the point (\(c_2\)) is said to be the local maxima.

Second Derivative Test For Inflection Point

In differential calculus, the concavity of the graph of a function is represented by the rate of change of the slope of a function. Thus, inflection points are points across which the rate of change of the slope of a curve or the graph of a function changes its sign, i.e., goes from negative to positive or positive to negative.

• Concave Upward : In the concave upward piece of the graph of a function, the slope of the graph goes on increasing, i.e., the rate of change of slope is positive.
• Concave Downward : In the concave downward piece, the slope of the graph goes on decreasing, i.e., the rate of change of slope is negative.

In differential calculus or differential geometry, the first order derivative of a function \(f^{‘}(x)\) at a point gives the rate of change of the function, i.e., the slope of the function at that point. Consequently, the second order derivatives of a function \(f^{“}(x)\) at a point, give the rate of change of the slope of the function at that point. Consequently, inflection points can also be considered as the points across which the second order derivative \(f^{“}(x)\) of the function \(f(x)\) goes from positive to negative or from negative to positive.

Thus, a point x = a on the graph of a function \(f(x)\) is said to be an inflection point if \(f^{“}(x) \) changes its sign across that point x=a and \(f^{“}(a)\)=0 or \(f^{“}(a)\) is undefined.

• Alternatively, a point x=a is an inflection point if \(f^{“}(a)\)=0 or \(f^{“}(a)\) is undefined and;
• \(f^{“}(x)\)<0 for x>a and \(f^{“}(x)\)>0 for x<a or
• \(f^{“}(x)\)>0 for x>a and \(f^{“}(x)\)<0 for x<a

Steps To Find Inflection Point With Second Derivative Test

The steps to find the inflection point with the second derivative test are as follows;

Step 1: Determine the first derivative i.e. \(\frac{d}{dx}f(x)\) of the given function i.e. f(x).

Step 2: Next, equate the received first derivative to zero i.e. \(\frac{d}{dx}f(x)=0\) and obtain the points.

Step 3: Now, specify the second derivative of the given function i.e. \(\frac{d^2}{dx^2}f(x)\).

Step 4: Lastly, substitute the points obtained in the second step in the second derivative and estimate the answer.

If both \(\frac{d}{dx}f(x)=0\text{ and }\frac{d^2}{dx^2}f(x)=0\), the point is said to be the inflection point.

Second Derivative Test for two Variables

Throughout the previous discussion, we saw how to check for local extrema and the point of inflection via the second derivative test for one variable. Let us understand the same for two variables.

If a function is given as u=f(x,y), then the steps to check for the local extrema and the point of inflection is as follows:

Step 1: First obtain the value of \(\frac{\partial u}{\partial x}\text{ and }\frac{\partial u}{\partial y}\).

Step 2: Solve for \(\frac{\partial u}{\partial x}=0\text{ and }\frac{\partial u}{\partial y}=0\), to obtain the stationary points.

Step 3: Next determine the values of \(r=\frac{\partial^2u}{\partial x^2},\ s=\frac{\partial^2u}{\partial x\partial y}\text{ and }t=\frac{\partial^2u}{\partial y^2}\).

Step 4: The condition for relative maxima, relative minima and saddle point are.

When \(rt-s^2>0\text{ and }r>0\) then the point of relative minima is obtained.

When \(rt-s^2>0\text{ and }r<0\) then the point of relative maxima is obtained.

When \(rt-s^2<0\), the point is neither maxima nor minima, i.e. it is the saddle point.

Importance of Second Derivative Test

Now that we know all about how to do the second derivative test through steps. Let us move ahead and understand the importance of the test.

• When the first derivative tells us whether the given function is increasing or decreasing, the second derivative tells us if the first derivative is increasing or decreasing.
• The second derivative test is also applied to locate the local maxima and local minima of a function with one variable, two variables and more under specific conditions.
• The test is also useful to obtain the inflection point or the saddle point.
• The test is also applied to determine the maximum or minimum value of a given function.
• The second derivative test is also useful in determining the nature of the curve (concave or convex) graphically.
• The second derivative test enables us to know if the given curve is concave up or concave down.

Solved Examples on Second Derivative Test

Well aware of the second derivative test and how to obtain the same for one and two variables. Let us go through some second derivative test practice problems

Solved Example 1: Obtain the critical points, local maxima and the local minima for the function\(f(x)=x^3-9x^2+15x+14\).

Solution: To obtain the critical points first obtain the first derivative of the function.

\(f^{\prime}(x)=3x^2-18x+15\)

Simplifying the expression we get;

\(f^{\prime}(x)=x^2-6x+5\)

Next, equate it to zero to get the critical point.

\(x^2-6x+5=0\)

Factorising the expression we get;

(x − 1)(x − 5) = 0

Thus the critical points are 1 and 5.

Now step towards finding the second derivative

\(f^{\prime\prime}(x)=6x-18\)

Substitute the value 1 and 5 in the above expression.

\(f^{\prime\prime}(1)=6-18=-12\)

Thus the given function has local maxima at x=1.

\(f^{\prime\prime}(5)=30-18=12\)

Therefore the given function has local minima at x=5.

Solved Example 2:Locate the inflection points for the given function; \(f(x)=x^4−24x^2+14\).

Solution: To start with obtaining the first and second derivatives of the given function.

\(f^{\prime}(x)=4x^3−48x\)

\(f^{\prime\prime}(x)=12x^2−48\)

For the inflection point, the f’’(x)=0

That is \(f^{\prime\prime}(x)=12x^2−48=0\)

On solving we get;

\(x^2=\frac{48}{12}=4\)

Thus x = 2 or x = -2

Let us check if x = 2 is an inflexion point or not. Thus take a point that is less than 2 and one that is greater than 2 and substitute it in the second derivative. Consider the points to be x=1 and x= 3

At x=1; \(f^{\prime\prime}(1)=12−48=-36\)

At x=3; \(f^{\prime\prime}(3)=108−48=60\)

We can say that \(\f”(x)\) changes sign across the point x = 2,which implies that the graph changes its nature; thus x=2 is an inflection point for the function f(x).

Similarly; for x = -2, take a point that is less than 2 and one that is greater than 2 and substitutes it in the second derivative. Consider the points to be x=-1 and x= -3

At x=-1; \(f^{\prime\prime}(-1)=12−48=-36\)

At x=-3; \(f^{\prime\prime}(-3)=108−48=60\)

Thus, here also \(\f”(x)\) changes sign across the point x = -2 which implies that the graph changes its nature; thus x=-2 is also an inflection point for the function f(x).

Solved Example 3: What are the possible situations that would rule out the use of the second derivative test for local extrema?

Solution: The possible situations that would rule out the use of the second derivative test for local extrema are;

• f’(x)=0 and f’’(x)=0 for the given expression or function.
• f’(x)=0 and f’’(x) does not exist for the given expression or function.
• f’(x) does not exist for the given expression or function.

Conclusion

In conclusion, the second derivative test is a powerful tool in calculus that helps you identify key points on a graph, like local maxima, minima, and inflection points. By understanding how the second derivative reveals the concavity of a function, you can determine where the graph changes direction. This is super important for solving problems efficiently, especially on exams like the SAT, ACT, AP, and more. Mastering this concept can really boost your math skills!