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zloy xaker [14]
2 years ago
11

Find the value of x for which the function y = x5-x³ + x² - 1 has a local minimum.

Mathematics
1 answer:
melisa1 [442]2 years ago
7 0

The given function presents a local minimum in the coordinates (0,-1).

<h3>Factoring</h3>

In math, factoring or factorization is used to write an algebraic expression in factors. There are some rules for factorization.  One of them is a factor out a common term for example: x²-x= x(x-1), where x is a common term.

For solving this question, the given equation should be rewritten  from the factoring.

x^5-x^3+x^2-1=\left(x+1\right)^2\left(x-1\right)\left(x^2-x+1\right)=0. Then, you have 3 equations.

From the Zero Factor Principle, you can write

Equation 1

(x+1)²=0

x+1=0

x= -1

Equation 2

x-1=0

x=1

Equation 3

x²-x+1=0

x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\ \\ x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{3}i}{2\cdot \:1}

From these points it is possible to plot a graph and you can see that the local minimum presents the coordinates (0,-1).

Read more about the factoring here:

brainly.com/question/11579257

#SPJ1

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Answer:

y = 3\, \sqrt{5} when x = 9.

Step-by-step explanation:

The question states that y is proportional to \sqrt{x}. In other words, there is a constant a (a \ne 0) such that y = a\, \sqrt{x} for all x \ge 0.

The question also states that y = x when x = 5. Make use of this equality to find the value of a.

Since x = 5 and y = x, it must be true that y = 5. Substitute x = 5\! and y = 5\! into the equation y = a\, \sqrt{x} and solve for a:

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\begin{aligned} y &= \sqrt{5} \, (\sqrt{9}) \\ &= \sqrt{5} \times 3 \\ &= 3\, \sqrt{5}\end{aligned}.

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