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f(x) the smallest integer a such that the Intermediate Value Theorem guarantees that has a zero on the interval (3, a); f(x) = - x ^ 2 + 6x - 8

          f(x) the smallest integer a such that the Intermediate Value Theorem guarantees that has a zero on the interval (3, a); f(x) = - x ^ 2 + 6x - 8
        

Added by Taylor C.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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f(x) the smallest integer a such that the Intermediate Value Theorem guarantees that has a zero on the interval (3, a); f(x) = - x ^ 2 + 6x - 8
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Transcript

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00:01 In this problem, we want to find the smallest integer a such that the intermediate value theorem guarantees that f of x has a 0 on the interval 0a, where f of x is defined as the parabola minus 8x squared plus 8x plus c.
00:22 So we want to find a 0 between 0 and a.
00:26 So let's first by finding our function zeros.
00:29 So our zeros correspond to locations where f of x is equal to 0.
00:45 This is the case when minus 8 x square plus 8x plus 6 is equal to 0.
00:57 Just divide everywhere by a factor 2, just to simplify, we have minus 4 x square plus 4 plus 3 equal to 0.
01:10 Now we can use our quadratic equation to solve for the 0.
01:13 We will have that our parabola will have zeros at x is equal to minus 4 plus or minus 4 square minus 4 times minus 4 times 3 all this divided by minus 4 times evaluating this expression we will have that a function has two zeros the first at minus one half and the second at three halves since we're only interested in the positive interval in zero and a we're only interested in this zero here so let's draw a graph to help us graphically what is going on let's draw a parabola with two zeros, one between minus one half and another zero at three halves.
02:51 And our parabola is going to be inverted because we have a negative number in front of our x squared term.
02:59 So our parabola is going to look something like this.
03:12 So between zero and three halves, f of x is positive...
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