Question

Given f(x) = x^3 - 5x² + 7x, find min/max by using 2nd Derivative Test only.

Answer 1

Explanation:

Suppose f(x) is a function of x that is twice differentiable at a stationary point x_0. Then, according to the Second Derivative Test, we have that:

Now, consider the following polynomial function:

$f(x)=x^3\text{ -5x}^2+7x$

the first derivate of this function is:

$f^{\prime}(x)=3x^2\text{ -10x + 7}$

and the second derivative would be:

$f^{\prime\prime}(x)=6x\text{ - 10}$

Now, to find the critical numbers we set the first derivative equal to 0:

$3x^2\text{ -10x + 7=0}$

the solutions to this equation are:

$x=\frac{7}{3}\text{ and x =1}$

To use the Second Derivative Test we evaluate the second derivative of f(x) at the above points (critical numbers):

$f^{\prime\prime}(\frac{7}{3})=6(\frac{7}{3})\text{ - 10=4 >0}$

and

$f^{\prime\prime}(1)=6(1)\text{ - 10= -4 <0}$

Applying the second derivative test, we have that according to these results, we have that f has a local minimum at x= 7/3 and f has a local maximum at x = 1.

Then, if we evaluate the function f(x) at the above critical points we obtain the coordinates for the maximum and minimum of the given function:

$f(\frac{7}{3})=(\frac{7}{3})^3\text{ -5\lparen}\frac{7}{3}\text{\rparen}^2+7(\frac{7}{3})=\frac{49}{27}$

and

$f(1)=(1)^3\text{ -5\lparen1\rparen}^2+7(1)=3$

we can conclude that the correct answer is:

Answer:

The maximum is the point:

$(1,3)$

The minimum is the point:

$(\frac{7}{3},\frac{49}{27})$