Complete factorization f(x)=x^4-x^3-5x^2-x-6
1 answer:
The solution to the above factorization problem is given as f′(x)=4x³−3x²−10x−1. See steps below.
<h3>What are the steps to the above answer?</h3>Step 1 - Take the derivative of both sides
f′(x)=d/dx(x^4−x^3−5x^2−x−6)
Step 2 - Use differentiation rule d/dx(f(x)±g(x))=d/dx(f(x))±d/dx(g(x))
f′(x)=d/dx(x4)−d/dx(x^3)−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−d/dx(x3)−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x2−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x^2−10x−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x^2−10x−1−dxd(6)
f′(x)=4x^3−3x^2−10x−1−0
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It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
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</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
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On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers