# calc

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Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].

• calc -

Hint:
For f(x) to have at most one root on [-2,2], it must be strictly increasing or decreasing on that interval.

So let's examine f'(x) on the interval [-2,2].

f(x)=x^3-15x+c
f'(x)=3x²-15
Absolute maximum on [-2,2] is when x=±2, f'(x)=4-15=-11
Absolute minimum on [-2,2] is when x=0, f'(x)=-15.

Therefore f'(x) is negative on [-2,2], therefore strictly decreasing. Under these circumstances, f(x) can have at most one zero (root).

• calc -

Thank you for your answer!!! My solution manual use Roll's theorem to solve it and I cannot understand it at all.
Your f'(2) and f'(-2) should equal -3 tho.

• calc -

yo, you plugged the numbers wrong bruh...
+ or - 2 when you plug it into f prime of x, you get -3 bruh not -11...

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