Show that the equation has 3 solutions in the interval ( - 4 , 4 ).

Question

zenmo · Answer

$$f(x) = x^3 - 15x + 1$$

zenmo · Answer

$$f(0) = (0)^3 - 15(x) + 1 = 1 > 0$$
$$f(-1) = (-1)^3-15(-1)+1 = -1 + 15 + 1 = 15 > 0$$
$$f(-2) = (-2)^3-15(-2)+1 = -8 + 30 + 1 = 23 > 0$$
Thus, by the Intermediate Value Theorem, the equation has 3 real solutions in the interval ( - 4, 4 ). I did not plug any numbers greater than 0 due to fact of obtaining a negative number and that the answer should be real numbers.

zenmo · Answer

I had this on my exam today, so I'm not sure if it is correct or not ...

sparrow2 · Answer

i think there are many was to show that

sparrow2 · Answer

ways

sparrow2 · Answer

when you wrote you didn't plug graeter then 0 because you get negative and the answer should be real number... i didn;t get this. so there aren't positive real numbers?

sparrow2 · Answer

as you wrote the question,i would do the following

sparrow2 · Answer

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