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Q. 60

Expert-verifiedFound in: Page 276

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

For each set of sign charts in Exercises *53–62*, sketch a possible graph of *f.*

The possible graph of *f *is

The given sign chart is

To sketch the possible graph of *f,* we will use theorem 3.6 and 3.10.

Theorem 3.6 states that the Derivative Measures Where a Function is Increasing or Decreasing, let *f* be a function that is differentiable on an interval *I*.

(a) If $f\text{'}$ is positive in the interior of *I*, then f is increasing on *I*.

(b) If $f\text{'}$ is negative in the interior of *I*, then f is decreasing on *I*.

(c) If $f\text{'}$ is zero in the interior of *I*, then f is constant on *I*.

Theorem 3.10 states that the Second Derivative Determines Concavity, suppose both *f* and $f\text{'}$ are differentiable on an interval *I*.

(a) If $f\text{'}\text{'}$ is positive on *I*, then* f* is concave up on *I*.

(b) If $f\text{'}\text{'}$ is negative on* I*, then *f* is concave down on *I*.

From the given chart, we conclude that

$f\text{'}$ is positive on the interval $\left(-\infty ,0\right)$ and negative on the interval $\left(0,\infty \right).$Thus, *f* will we increase on the positve intervals and decrease on the negative intervals.

$f\text{'}\text{'}$ is positive on the intervals $\left(-\infty ,-1\right)\mathrm{and}\left(1,\infty \right)$ and negative on the interval $\left(-1,1\right).$ Thus, *f *will be concave up on the positive intervals and concave down on the negative intervals.

The graph has roots at $x=-1,$ a local minimum at $x=0,$ and inflection points at $x=-1and1.$

The graph is

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