Make use of the graph off(x)=x2to find a number&nbsp;(δ)&nbsp;such thatif&nbsp;|x−1|&lt;(δ)&nbsp;then&nbsp;|x2−1|&lt;12(Round your response to the nearest three decimal places.)(δ)=

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Make use of the graph off(x)=x2to find a number&amp;nbsp;(δ)&amp;nbsp;such thatif&amp;nbsp;|x−1|&amp;lt;(δ)&amp;nbsp;then&amp;nbsp;|x2−1|&amp;lt;12(Round your response to the nearest three decimal places.)(δ)=

unessodopunsep · Accepted Answer

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nick1337 · Accepted Answer

Step 1: Find the expression for |x2−1|.Since f(x)=x2, we can rewrite |x2−1| as |f(x)−f(1)|.Step 2: Determine the conditions for |f(x)−f(1)|&amp;lt;12.Using the definition of the absolute value, we have −12&amp;lt;f(x)−f(1)&amp;lt;12.Step 3: Express f(x)−f(1) in terms of x.Substituting f(x)=x2 and f(1)=12=1, we get −12&amp;lt;x2−1&amp;lt;12.Step 4: Solve the inequality.Rearranging the inequality, we have −12&amp;lt;x2−1&amp;lt;12.To solve this inequality, we can analyze it by splitting it into two separate inequalities:−12&amp;lt;x2−1  ... (1)x2−1&amp;lt;12  ... (2)Now, let&#039;s solve each inequality individually:Solving inequality (1):−12&amp;lt;x2−1Adding 1 to both sides:12&amp;lt;x2Taking the square root:12&amp;lt;|x|Since we&#039;re interested in the distance between x and 1, we can rewrite this as:|x−1|&amp;gt;12Simplifying the square root:|x−1|&amp;gt;22  ... (3)Solving inequality (2):x2−1&amp;lt;12Adding 1 to both sides:x2&amp;lt;32Taking the square root:|x|&amp;lt;32Again, since we&#039;re interested in the distance between x and 1:|x−1|&amp;lt;32  ... (4)Step 5: Determine the maximum value for δ.To satisfy both conditions (3) and (4), we need to choose the smaller value between 22 and 32.Therefore, δ=22.Hence, if |x−1|&amp;lt;δ, where δ=22, then |x2−1|&amp;lt;12.

madeleinejames20 · Accepted Answer

Result:0.293Solution:To find the value of δ such that if |x−1|&amp;lt;δ, then |x2−1|&amp;lt;1/2, we can utilize the graph of f(x)=x2.Let&#039;s start by considering the inequality |x2−1|&amp;lt;1/2. This means that the distance between x2 and 1 should be less than 1/2. We can rewrite this inequality as −1/2&amp;lt;x2−1&amp;lt;1/2.Now, let&#039;s focus on the right-hand side inequality: x2−1&amp;lt;1/2. Adding 1 to both sides gives us x2&amp;lt;1+1/2, which simplifies to x2&amp;lt;3/2.To find δ, we need to determine the interval around x = 1 such that all values of x within this interval satisfy the inequality x2&amp;lt;3/2. Since the graph of f(x)=x2 is symmetric around x = 0 and opens upward, we can consider the distance from x = 1 to the point where x2=3/2.To find this point, we solve the equation x2=3/2:x2=32Taking the square root of both sides, we get:x=±32Since we are interested in the interval around x = 1, we take the positive square root:x=32To find δ, we take the absolute value of the difference between 1 and the x-coordinate of this point:δ=|1−32|Now, let&#039;s compute the value of δ:δ=|1−32|≈0.293Therefore, the value of δ that satisfies the given condition is approximately 0.293.

Eliza Beth13 · Accepted Answer

We know that the function f(x)=x2 is a quadratic function with a vertex at (0, 0) and it opens upward. The graph of f(x)=x2 is a U-shaped curve.To find δ, we need to determine the interval around x = 1 where |x2−1|&amp;lt;1/2 holds true.Let&#039;s start by examining the expression |x2−1|&amp;lt;1/2:|x2−1|&amp;lt;12We can rewrite the inequality as:−12&amp;lt;x2−1&amp;lt;12Next, let&#039;s add 1 to all parts of the inequality:−12+1&amp;lt;x2−1+1&amp;lt;12+1Simplifying further:12&amp;lt;x2&amp;lt;32Now, taking the square root of all parts of the inequality, we have:12&amp;lt;|x|&amp;lt;32Since we are interested in the interval around x = 1, we can subtract 1 from each part:12−1&amp;lt;|x−1|&amp;lt;32−1To ensure that |x−1|&amp;lt;δ, we need to find a δ that satisfies the above inequality.Hence, we can choose:δ=min(12−1,32−1)Therefore, the value of δ that satisfies the given condition is:δ=12−1

Use the given graph off(x) = x^2to find a number (delta) such thatif |x - 1| < (delta) then |x^2-1| < \frac{1}{2}(Round your answer down to three decimal places.)(delta) =1210500781.jpg

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