Where r the turning points of x^2(x-4)^7(x+3)^6

Mathematics

1 answer:

Tomtit [17]3 years ago

8 0

Answer:

There are 5 turning points at x = -3, -1.11, 0, 1.44 and 4.

Step-by-step explanation:

The product rule for 3 terms is

F' (x) = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)

F(x) = x^2(x - 4)^7 (x + 3)^6

F' (x) = 2x(x-4)^7(x+3)^6 + x^2 7(x - 4)^6 (x + 3)^6 + x^2(x-4)^7 6(x+3)^5 = 0 at the turning points.

That is some equation!!

By observation, 3 roots are x = 0 , x = 4 and x = -3 so there are turning points at these values of x.

I'll use software to graph it to find any more real solutions - they are -1.11 and 1.44 to nearest hundredth.

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Help again...its for math offering 15 points

Anettt [7]

Answer:

$\tan \dfrac{\theta}2 = \dfrac{1}2$

Step by step Explanation:

$\sin \theta = \dfrac{\text{Perpendicular} }{\text{Hypotenuse}} = \dfrac{12}{15}\\\\\\\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}= \dfrac{9}{15}\\\\\text{Now,}\\\\\tan \dfrac{\theta}2 = \dfrac{\sin \tfrac{\theta}2}{\cos \tfrac{\theta}2}\\\\\\~~~~~~~~=\dfrac{2\cos \tfrac{\theta}2 \sin \tfrac{\theta}2 }{2\cos^2 \tfrac{\theta}2}~~~~~~;\left[\text{Multiply by}~ 2\cos\tfrac{\theta}2 \right]$

$=\dfrac{\sin \theta}{1+ \cos \theta}~~~~~~~~~~~~;[2 \sin x \cos x = \sin 2x ~ \text{and}~ 2\cos^2 x =1+\cos 2x]\\\\\\=\dfrac{\tfrac{12}{15}}{1+ \tfrac{9}{15}}\\\\\\=\dfrac{\tfrac{12}{15}}{\tfrac{24}{15}}\\\\\\=\dfrac{12}{15}\times \dfrac{15}{24}\\\\\\=\dfrac{12}{24}\\\\\\=\dfrac{1}2$