f(x) = (x^3)-7(x^2)+25x+8
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My attempt:
I can easily prove by the IVT that an arbitrary interval [-1,1] proves i have AT LEAST one root. f(-1)=-25, f(1) = 27. So, f(-1) < 0 < f(1), proves at least one c within the interval.
Now i need to restrict the the root to no more than 1. I figure i could prove that the function is either always increasing or decreasing.
Since the derivtive is ->
f'(x) = 3x^2-14x+25
3x^2 = increasing
-14x = increasing/decreasing
25 = increasing
So since not all terms are always increasing( or decreasing), i can't prove that the roots are restricted to exactly one, even though a graphing calculator proves exactly 1 root.
--------------------------------------…
My attempt:
I can easily prove by the IVT that an arbitrary interval [-1,1] proves i have AT LEAST one root. f(-1)=-25, f(1) = 27. So, f(-1) < 0 < f(1), proves at least one c within the interval.
Now i need to restrict the the root to no more than 1. I figure i could prove that the function is either always increasing or decreasing.
Since the derivtive is ->
f'(x) = 3x^2-14x+25
3x^2 = increasing
-14x = increasing/decreasing
25 = increasing
So since not all terms are always increasing( or decreasing), i can't prove that the roots are restricted to exactly one, even though a graphing calculator proves exactly 1 root.
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The discriminant of the derivative is (-14)^2 - 4(3)(25) = 196-300 = -104. The negative value means the quadratic has no real roots, so is positive everywhere. (It has a value of +25 at x=0).