Question

use the first derivative test to classify the relative extrema

Answer 1

Answer:

Local maximum: (-1/5, -10)

Local minimum: (1/5, 10)

Explanation:

Given the below function;

$f(x)=\frac{25x^2+1}{2}$

To determine the relative maxima of the above function using the First Derivative Test, we'll follow the below steps;

Step 1: Determine the derivative of f(x) using the Quotient rule;

Let u = 25x^2 + 1

u' = 50x

Let v = x

v' = 1

$\begin{gathered} f^{\prime}(x)=\frac{v\cdot u^{\prime}-u\cdot v^{\prime}}{v^2}_{} \\ f^{\prime}(x)=\frac{x(50x)-(25x^2+1)(1)}{x^2} \\ f^{\prime}(x)=\frac{25x^2-1}{x^2} \end{gathered}$

Step 2: Determine the critical points by equating f(x) to zero and solving for x;

$\begin{gathered} \frac{25x^2-1}{x^2}=0 \\ 25x^2-1=0 \\ 25x^2=1 \\ x^2=\frac{1}{25} \\ x=\pm\sqrt[]{\frac{1}{25}} \\ x=\pm\frac{1}{5} \end{gathered}$

Step 3: Since the critical points are at -1/5 and 1/5, we'll check if the first derivative changes signs around these points by picking numbers from different intervals as seen below;

From negative infinity to -1/5, let's pick -1;

$f^{\prime}(-1)=\frac{25(-1)^2-1}{(-1)^2}=\frac{25-1}{1}=24$

From -1/5 to 1/5, let's pick 0.1;

$f^{\prime}(0.1)=\frac{25(0.1)^2-1}{(0.1)^2}=-75$

From 1/5 to infinity; let's pick 1;

$f^{\prime}(1)=\frac{25(1)^2-1}{(1)^2}=24$

We can see from the above that the first derivative changes from positive to negative around x = -1/5, it signifies that this critical point is a local maximum.

It also changes from negative to positive around x = 1/5, it signifies that this critical point is a local minimum.

Let's go ahead and determine the actual points at which the function has the local minimum and maximum by evaluating f(x) at the critical points;

$\begin{gathered} f(-\frac{1}{5})=\frac{25(-\frac{1}{5})^2+1}{(-\frac{1}{5})}=-10 \\ f(\frac{1}{5})=\frac{25(\frac{1}{5})^2+1}{(\frac{1}{5})}=10 \end{gathered}$

Therefore, the function f(x) has a local maximum at (-1/5, -10) and a local minimum at (1/5, 10)