Maxima and minima with second derivative – Examples with answers

To find the maximum point, we have to start by finding all the stationary points. For this, we obtain the derivative of the function

$latex f(x)=-x^2+2x$

$latex f'(x)=-2x+2$

Now, we form an equation with the derivative and find the values of x:

$latex -2x+2=0$

$latex -2x=-2$

$latex x=1$

The value found corresponds to a stationary point. To check if the point is a maximum point, we use the second derivative:

$latex f^{\prime \prime}(x)=-2$

Since $latex f^{\prime \prime}(x)<0$, the found point corresponds to a maximum point.

Using $latex x=1$ in the function, we find the y-coordinate of the maximum point:

$latex y=-x^2+2x$

$latex y=-(1)^2+2(1)$

$latex y=1$

The maximum point has the coordinates (1, 1).

We start by finding the derivative of the function since we are going to use it to find the stationary points:

$latex f(x)=x^2+2x-1$

$latex f'(x)=2x+2$

We form an equation with the derivative and solve for x:

$latex 2x+2=0$

$latex 2x=-2$

$latex x=-1$

Now, we use the second derivative to check if the point is a minimum point.

$latex f^{\prime \prime}(x)=2$

Since we have $latex f^{\prime \prime}(x)>0$, the point found corresponds to a minimum point.

Finally, we use $latex x=-1$ in the function to determine the y-coordinate of the point:

$latex y=x^2+2x-1$

$latex y=(-1)^2+2(-1)-1$

$latex y=-2$

The minimum point of the function is (-1, -2).

We start by finding the derivative of the function:

$latex f(x)=-x^3+3x+2$

$latex f'(x)=-3x^2+3$

Forming an equation with the derivative, we can solve for x to find the stationary points:

$latex -3x^2+3=0$

$latex -3x^2=-3$

$latex x=\sqrt{3}$

$latex x=1~~$ or $latex ~~x=-1$

Using the second derivative, we can determine the nature of the points found:

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

If $latex x=1$, we have $latex f^{\prime \prime}(x)<0$, and if $latex x=-1$, we have $latex f^{\prime \prime }(x)>0$. Then, $latex x=1$ is the maximum point.

We use $latex x=1$ to find the y-coordinate of the maximum point:

$latex y=-x^3+3x+2$

$latex y=-(1)^3+3(1)+2$

$latex y=4$

The maximum point has the coordinates (1, 4).

The derivative of the function is as follows:

$latex f(x)=x^3-3x-2$

$latex f'(x)=3x^2-3$

Now, we use the derivative to form an equation and solve for x:

$latex 3x^2-3=0$

$latex x=\sqrt{1}$

$latex x=\pm 1$

$latex x=1~~$ or $latex ~~x=-1$

We have two stationary points. Therefore, we have to use the second derivative to determine the nature of these points:

$latex f^{\prime \prime}(x)=6x$

When $latex x=1$, we have $latex f^{\prime \prime}(x)>0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)< 0$. Thus, $latex x=1$ is the minimum point.

We find the y-coordinate of the minimum point using $latex x=1$ in the function:

$latex y=x^3-3x-2$

$latex y=(1)^3-3(1)-2$

$latex y=-4$

The minimum point of the function is (1, -4).

The derivative of the function is:

$latex f(x)=2x^3-6x+5$

$latex f'(x)=6x^2-6$

We can find the x-values of the stationary points by forming an equation with the derivative:

$latex 6x^2-6=0$

$latex 6x^2=6$

$latex x^2=\sqrt{1}$

$latex x=1~~$ or $latex ~~x=-1$

Now, we use the second derivative to determine the nature of these points:

$latex f^{\prime \prime}(x)=12x$

When $latex x=1$, we have $latex f^{\prime \prime}(x)>0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)< 0$, so $latex x=-1$ is the maximum point.

We find the y-coordinate of the maximum point as follows:

$latex y=2x^3-6x$

$latex y=2(-1)^3-6(-1)$

$latex y=4$

The maximum point has the coordinates (-1, 4).

We find the derivative of the function as follows:

$latex f(x)=2x^3-6x-4$

$latex f'(x)=6x^2-6$

We can find the stationary points by forming an equation with the derivative and solving for x:

$latex 6x^2-6=0$

$latex x=\sqrt{1}$

$latex x=\pm 1$

$latex x=1~~$ or $latex ~~x=-1$

We have two stationary points, so we use the second derivative to determine their nature:

$latex f^{\prime \prime}(x)=12x$

When $latex x=1$, we have $latex f^{\prime \prime}(x)>0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)< 0$. This means that $latex x=1$ is the minimum point.

Finally, we find the y-coordinate of the point:

$latex y=2x^3-6x-4$

$latex y=2(1)^3-6(1)-4$

$latex y=-8$

The minimum point of the function is (1, -8).

We start by finding the derivative of the function:

$latex f(x)=\frac{1}{3}x^3+x^2-3x$

$latex f'(x)=x^2+2x-3$

Now, let’s form an equation with the derivative and solve by factoring:

$latex x^2+2x-3=0$

$latex (x+3)(x-1)=0$

$latex x=-3~~$ or $latex ~~x=1$

We found two stationary points, so we can use the second derivative to determine the nature of these points:

$latex f^{\prime \prime}(x)=2x+2$

When $latex x=-3$, we have $latex f^{\prime \prime}(x)<0$ and when $latex x=1$, we have $latex f^{\prime \prime}(x)> 0$. Thus, the maximum point is at $latex x=-3$.

Finally, we find the y-coordinate of the point using $latex x=-3$ in the function:

$latex f(x)=\frac{1}{3}x^3+x^2-3x$

$latex f(x)=\frac{1}{3}(-3)^3+(-3)^2-3(-3)$

$latex y=9$

The maximum point has the coordinates (-3, 9).

We start by finding the derivative of the function and forming an equation to get the stationary points:

$latex f(x)=-\frac{1}{3}x^3+x^2+3x$

$latex f'(x)=-x^2+2x+3$

$latex -x^2+2x+3=0$

$latex -(x-3)(x+1)=0$

$latex x=3~~$ or $latex ~~x=-1$

Now, we use the second derivative

$latex f^{\prime \prime}(x)=-2x+2$

When $latex x=3$, we have $latex f^{\prime \prime}(x)<0$ and when $latex x=-1$, we have $latex f^{\prime \prime}(x)> 0$. This means that $latex x=-1$ is the minimum point.

Finally, we determine the y-coordinate of the minimum point:

$latex y=-\frac{1}{3}x^3+x^2+3x$

$latex y=-\frac{1}{3}(-1)^3+(-1)^2+3(-1)$

$latex y=-\frac{5}{3}$

The minimum point of the function is $latex (-1, ~-\frac{5}{3})$.

We can solve this problem by finding a function of the area in terms of x. For this, we observe the following:

$latex AB+BC=6$

$latex x+BC=6$

$latex BC=6-x$

Then, we can use the triangle area formula to get:

$latex A=\frac{1}{2}\times BC \times AB$

$latex A=\frac{1}{2}(6-x)x$

$latex A=3x-\frac{x^2}{2}$

Now, we have to find the derivative to form an equation and find the stationary points:

$latex A(x)=3x-\frac{x^2}{2}$

$latex A'(x)=3-x$

$latex 3-x=0$

$latex x=3$

We check whether this point is a maximum point using the second derivative:

$latex A^{\prime \prime}(x)=-1$

Since $latex f^{\prime \prime}(x)<0$, we know that point is a maximum. Then, we use $latex x=3$ to find the maximum area:

$latex A=3x-\frac{x^2}{2}$

$latex A=3(3)-\frac{(3)^2}{2}$

$latex A=\frac{9}{2}=4.5$

The maximum area of the triangle is 4.5 cm².

Similar to the previous exercise, we have to find a function for the area of the triangle using the given information:

$latex A=\frac{1}{2}\times BC \times AB$

$latex A=\frac{1}{2}(x+2)x$

$latex A=\frac{1}{2}x^2+x$

Now, we find the derivative to form an equation and find the stationary points:

$latex A(x)=\frac{1}{2}x^2+x$

$latex A'(x)=x+1$

$latex x+1=0$

$latex x=-1$

To check if the point is a minimum, we are going to use the second derivative:

$latex A^{\prime \prime}(x)=1$

We have $latex f^{\prime \prime}(x)>0$, so we confirm that the point is a minimum. Then, we use $latex x=-1$ to find the minimum area:

$latex A=\frac{1}{2}x^2+x$

$latex A=\frac{1}{2}(-1)^2-1$

$latex A=-\frac{1}{2}=0.5$

The minimum area of the triangle is 0.5 cm².