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Find all fixed point(s) of the given function φ(x)=x²−4x−6 within the specified interval. (a) φ(x) on [0,10]

User Alokj
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Final answer:

To find the fixed point(s) of the function φ(x), set φ(x) = x and solve the resulting quadratic equation. The only fixed point within the interval [0, 10] is at x = 6, as the other solution is outside of the given interval.

Step-by-step explanation:

To find the fixed point(s) of the function φ(x) = x² - 4x - 6 within the interval [0,10], we need to solve the equation where φ(x) = x. This is the definition of a fixed point: a point x at which φ(x) = x.


So, set up the equation:

x = x² - 4x - 6

Then rearrange the terms to set the equation to zero:

0 = x² - 4x - 6 - x

0 = x² - 5x - 6

This is now a quadratic equation in the form ax² + bx + c = 0, where a=1, b=-5, and c=-6.

To solve for x, we will use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Substituting a, b and c gives us:

x = (5 ± √(25 - 4(1)(-6))) / (2)

x = (5 ± √(25 + 24)) / 2

x = (5 ± √(49)) / 2

x = (5 ± 7) / 2

This gives us two potential solutions: x = (5 + 7) / 2 = 6 and x = (5 - 7) / 2 = -1

However, remember we are looking for fixed point(s) within the interval [0, 10]. The value x = -1 is not within this interval, so we ignore it.

Thus, the only fixed point of the function φ(x) within the interval [0, 10] is at x = 6.

User Luis Guzman
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