The value of -1 is a lower bound for the zeros of the function show below.

1 answer:

7 0

To perform this check, you must use the following theorem: $a$ is a lower bound for the zeroes of $f(x)$ if, when divide $f(x)$ by $(x-a)$ , the quotient and the remainter alternate signs.

Since the long division yields the result

$\dfrac{x^4+x^3-11x^2-9x+18}{x+1}=x^3-11x+2+\dfrac{16}{x+1}$

You can see that the signs don't alternate (the last two terms are positive). So, -1 is not a lower bound.

In fact, the actual roots of the polynomial are -3, -2, 1, 3, so as you can see there are roots smaller than -1.

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Checking the Mean Value Theorem:<br><br>Number 3

mrs_skeptik [129]

$\bf f(x)=x+\cfrac{1}{x}\qquad \left[\frac{1}{2},2 \right]\\\\
-----------------------------\\\\
\cfrac{df}{dx}=1+\left(-1x^{-2} \right)\implies \cfrac{df}{dx}=1-\cfrac{1}{x^2}
\\\\\\
f'(c)=1-\cfrac{1}{c^2}\quad \quad 1-\cfrac{1}{c^2}=\cfrac{f(2)-f\left( \frac{1}{2} \right)}{2-\frac{1}{2}}
\\\\\\
1-\cfrac{1}{c^2}=\cfrac{\frac{5}{2}-\frac{5}{2}}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=0
\\\\\\
1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm\sqrt{1}\implies c=\pm 1$

there's a quick graph below of the bounds and the tangent at "c"

not happening -2 or 2 will have a tangent parallel to a,b, needless to say -2 is out of the range [a,b] anyway, so the only value is really 1, on the positive 1st quadrant