The derivative of a function f is given by f^(')(x)=ln |x|*x. On which interval is the graph of f concave up? Use a graphing calculator. Choose 1 answer: (A) -1 1 (B) x > 0 (C) x 0.368 (E) All real numbers

Calculate First Derivative: To determine where the graph of f is concave up, we need to find the second derivative, f''(x), and look for where it is positive. f'(x) = ln|x|*x f''(x) = d/dx(ln|x|*x)Apply Product Rule: Use the product rule for differentiation: (uv)' = u'v + uv'. Let u = ln|x| and v = x. Then u' = 1/x and v' = 1. f''(x) = (1/x)*x + ln|x|*1Simplify Second Derivative: Simplify the second derivative: f''(x) = 1 + ln|x|Determine Positive Values: Find where f''(x) is positive: 1 + ln|x| > 0 ln|x| > -1Solve for x (x > 0): Solve for x in the inequality ln|x| > -1. For x > 0: ln(x) > -1 For x -1Solve for x (x 0: e^(ln(x)) > e^(-1), so x > 1/e For x e^(-1), so -x > 1/e, which means x < -1/e

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The derivative of a function
f is given by
f^(')(x)=ln |x|*x.
On which interval is the graph of
f concave up?
Use a graphing calculator.
Choose 1 answer:
(A)
-1 < x < 0 and
x > 1
(B)
x > 0
(C)
x < 0
(D)
x < -0.368 and
x > 0.368
(E) All real numbers

The derivative of a function $f$ is given by $f^{\prime}(x)=\ln |x| \cdot x$ . $\newline$ On which interval is the graph of $f$ concave up? $\newline$ Use a graphing calculator. $\newline$ Choose $1$ answer: $\newline$ (A) $-1<x<0$ ="" and="" $="" x="">1$ $\newline$ (B) x>0 $\newline$ (C) x<0 $\newline$ (D) x<-0.368 and x>0.368 $\newline$ (E) All real numbers

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Q. The derivative of a function

f

is given by

f^{\prime}(x)=\ln |x| \cdot x

\newline

On which interval is the graph of

f

concave up?

\newline

Use a graphing calculator.

\newline

Choose

1

answer:

\newline

(A)

-1<x<0

and

x>1

\newline

(B)

x>0

\newline

(C)

x<0

\newline

(D)

x<-0.368

and

x>0.368

\newline

(E) All real numbers

Calculate First Derivative: To determine where the graph of $f$ is concave up, we need to find the second derivative, $f''(x)$ , and look for where it is positive. $\newline$ $f'(x) = \ln|x|\cdot x$ $\newline$ $f''(x) = \frac{d}{dx}(\ln|x|\cdot x)$
Apply Product Rule: Use the product rule for differentiation: $uv$ ' = u'v + uv'. Let $u = \ln|x|$ and $v = x$ . Then $u' = \frac{1}{x}$ and $v' = 1$ . $f''(x) = \left(\frac{1}{x}\right)x + \ln|x|\cdot 1$
Simplify Second Derivative: Simplify the second derivative: $f''(x) = 1 + \ln|x|$
Determine Positive Values: Find where $f''(x)$ is positive: 1 + \ln|x| > 0 \ln|x| > -1
Solve for x (x > 0): Solve for x in the inequality \ln|x| > -1. For x > 0: \ln(x) > -1 For x < 0: \ln(-x) > -1
Solve for x x < 0: Exponentiate both sides of the inequalities to solve for x. $\newline$ For x > 0: e^{\ln(x)} > e^{-1}, so x > \frac{1}{e} $\newline$ For x < 0: e^{\ln(-x)} > e^{-1}, so -x > \frac{1}{e}, which means x < -\frac{1}{e}