Question

What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?

Answer 1

Hello,

$y=x^5-10x^3+9x\\ y'=5x^4-30x^2+9=0\\ \Delta=30^2-4*5*9=720=(12\sqrt{5})^2\\\\ x_{1} =- \sqrt{3- \frac{6 \sqrt{5} }{5} } =-0.5627774222...\\ y_{1}= -3.3390295...\\\\ x_{2} =- \sqrt{3+ \frac{6 \sqrt{5} }{5} } =-2.3839....\\ y_{2}= 37.0304482...\\\\ x_{3} = \sqrt{3- \frac{6 \sqrt{5} }{5} } =0.5627774222...\\ y_{3}= 3.3390295...\\\\ x_{4} = \sqrt{3+ \frac{6 \sqrt{5} }{5} } =2.3839....\\ y_{4}= -37.0304482...$

Answer 2

Remember that the maximum and minimum values of a function could occur when the derivative of the function is equal to zero or is undefined, or they could also occur at the endpoints of the interval.

The first thing we are going to do is find the derivative of our function:

$f(x)=x^5-10x^3+9x$

$f'(x)=\frac{dy}{dx} x^5-\frac{dy}{dx} 10x^3+\frac{dy}{dx} 9x$

$f'(x)=5x^4-30x^2+9$

Next, we are going to set our derivative equal to zero and solve for $x$:

$5x^4-30x^2+9=0$

Solving a polynomial of degree 4 is problematic, but we can substitute the variable here to get a quadratic equation so we can use the quadratic formula. let $x^2=u$

$5u^2-30u+9=0$

Using the quadratic formula we get that:

$u=3+\frac{6\sqrt{5}}{5} , u=3-\frac{6\sqrt{5}}{5}$

But remember that $u=x^2$, so we need to substitute back $x^2$:

$x^2=3+\frac{6\sqrt{5}}{5} , x^2=3-\frac{6\sqrt{5}}{5}$

Now we can solve for $x$ to get:

$x=\sqrt{3+\frac{6\sqrt{5}}{5}} , x=-\sqrt{3+\frac{6\sqrt{5}}{5}} , x=\sqrt{3-\frac{6\sqrt{5}}{5}} ,x=-\sqrt{3-\frac{6\sqrt{5}}{5}}$

Since the problem is asking us for the proximate values of the maximum and minimum, we can round our solutions to the nearest hundredth:

$x=2.38,x=-2.38,x=0.56,x=-0.56$

Now, we just need to evaluate our original function at those points, but remember that the maximum and minimum points could occur at the endpoints, so we need to evaluate our function at $x=3$ and $x=-3$ as well.

$f(x)=x^5-10x^3+9x$

for $x=2.38$, $f(2.38)=(2.38)^5-10(2.38)^3+9(2.38)=-37.03$

for $x=-2.38$, $f(-2.38)=(-2.38)^5-10(-2.38)^3+9(-2.38)=37.03$

for $x=0.56$, $f(0.56)=(0.56)^5-10(0.56)^3+9(0.56)=3.34$

for $x=-0.56$, $f(-0.56)=(-0.56)^5-10(-0.56)^3+9(-0.56)=-3.34$

for $x=3$, $f(3)=(3)^5-10(3)^3+9(3)=0$

for $x=-3$, $f(-3)=(-3)^5-10(-3)^3+9(-3)=0$

Notice that the maximum value of the function is 3.34, and it occur when $x=0.56$; the minimum value is -37.03, and it occur when $x=2.38$

We can conclude the minimum point of the function is (2.38, -37.03), and the maximum point is (0.56, 3.34).