Question

What is the maximum local value of the function?
g(x)=x^4-5x^2+4
i tried solving the function when g(x)=0 and got x=1 or -1. I dont think those are right.

Answer 1

please you have to start from the first derivative of your function \(g(x)\)

Answer 2

@Michele_Laino Explain please? I'm new to this and want to see examples to help :)

Answer 3

the first derivative of your function is:
\[g'\left( x \right) = 4{x^3} - 10x\]

Answer 4

How did you get that? @Michele_Laino

Answer 5

I have applied the rules of differential calculus

Answer 6

for example the first derivative of this function: \(y=k x^n\) is:
\(y'=knx^{n-1}\)

Answer 7

where \(k\) is a real number and \(n \) is a natural number

Answer 8

Oh okay, i see now. So do you take that and set it to 0?

Answer 9

we have the subsequent theorem:
"A function \(f(x)\) is a not decreasing function at point \(x\) such that: \(f'\left( x \right) \geqslant 0\)"

Answer 10

In other words, in order to get the point of local maximum for \(f(x9\) we have to solve this inequality:
\[4{x^3} - 10x \geqslant 0\]
or:
\[2x\left( {2{x^2} - 5} \right) \geqslant 0\]

Answer 11

oops.. for \(f(x)\)*...

Answer 12

whch is easier to solve for?

Answer 13

the inequality above, is equivalent to these systems of inequalities:
\[\left\{ {\begin{array}{*{20}{c}}
{x \geqslant 0} \\
{2{x^2} - 5 \geqslant 0}
\end{array}} \right. \cup \left\{ {\begin{array}{*{20}{c}}
{x \leqslant 0} \\
{2{x^2} - 5 \leqslant 0}
\end{array}} \right.\]

Answer 14

as you can see, between them I wrote the symbol of "union", since we have to make the union of the solutions of both systems above

Answer 15

okay, so do we go on to solve?

Answer 16

for example I try to solve the first system:

\[\left\{ {\begin{array}{*{20}{c}}
{x \geqslant 0} \\
{2{x^2} - 5 \geqslant 0}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x \geqslant 0} \\
{x \geqslant \sqrt {\frac{5}{2}} ,\;x \leqslant - \sqrt {\frac{5}{2}} }
\end{array}} \right.\]
so its solution is given by the subsequent interval of the real line:
\[x \geqslant \sqrt {\frac{5}{2}} \]

Answer 17

please try to solve the second system of inequalities above

Answer 18

1.5811?

Answer 19

yes! It is correct, nevertheless, please keep in mind that our interest is in finding the solution of the second system of inequalities

Answer 20

will it be -1.5811?

Answer 21

no, I'm sorry
I try to solve the second system:
\[\left\{ {\begin{array}{*{20}{c}}
{x \leqslant 0} \\
{2{x^2} - 5 \leqslant 0}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x \leqslant 0} \\
{ - \sqrt {\frac{5}{2}} \leqslant x \leqslant \sqrt {\frac{5}{2}} }
\end{array}} \right.\]

Answer 22

so, the solution of the second system is the subsequent interval of the real line:
\[ - \sqrt {\frac{5}{2}} \leqslant x \leqslant 0\]

Answer 23

-1.581<x<0

Answer 24

Now, I make the set union of both solutions, so i can say that our function is a nondecreasing function for all point \(x\) which belong to this interval of the real line: