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If possible, choose k so that the following function is continuous on any interval:

f(x)= (5x^4-20x^3)/(x-4) , x≠4
f(x)= K , x=4

k=?,

Respuesta :

Mindaka
We need to cancel out the discontinuity in x = 4

In order to do that, factorize the numerator:
5x⁴ - 20x³ = 5x³(x-4)

This way, your function will be:
f(x) = 5x³(x-4) / (x-4)

and the two parentheses cancel out, leaving
f(x) = 5x³ 

which at x = 4 gives:
f(4) = 5·4³ = 5 · 64 = 320

Therefore K = 320.

Using continuity concepts, it is found that K = 320 makes the function continuous.

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A function f(x) is said to be continuous at x = a if:

[tex]\lim_{x \rightarrow a^{-}} = \lim_{x \rightarrow a^{+}} = f(a)[/tex]

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The piece-wise function given is:

[tex]f(x) = \frac{5x^4 - 20x^3}{x-4}, x \neq 4[/tex]

[tex]f(x) = K, x = 4[/tex]

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The definition for x different of 4 can be simplified, as follows:

[tex]f(x) = \frac{5x^4 - 20x^3}{x-4} = \frac{5x^3(x - 4)}{x - 4} = 5x^3[/tex]

Thus, the simplified definition is:

[tex]f(x) = 5x^3, x \neq 4[/tex]

[tex]f(x) = K, x = 4[/tex]

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For the lateral limits, x is close to but not exactly 4, so the first definition is used.

[tex]\lim_{x \rightarrow 4^-} = \lim_{x \rightarrow 4} 5x^3 = 5(4)^3 = 5 \times 64 = 320[/tex]

[tex]\lim_{x \rightarrow 4^+} = \lim_{x \rightarrow 4} 5x^3 = 5(4)^3 = 5 \times 64 = 320[/tex]

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For continuity, it is needed that:

[tex]f(4) = 320[/tex]

Thus:

[tex]K = 320[/tex] makes the function continuous.

A similar problem is given at https://brainly.com/question/14755764

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