The derivative of a function g is given by g^(')(x)=x^(3)-2x^(2)-3x+4. On which intervals is the graph of g increasing? Use a graphing calculator. Choose 1 answer: (A) x = 2.562 (C) x = 1.869 (D) -0.535 <= x <= 1.869 (E) All real numbers

Find Critical Points: Find the critical points of g'(x) by setting the derivative equal to zero and solving for x. g'(x) = x^3 - 2x^2 - 3x + 4 = 0Use Graphing Calculator: Use a graphing calculator to find the roots of the equation. Roots are approximately x = -0.535, x = 1, and x = 1.869.Determine Sign of g'(x): Determine the sign of g'(x) on the intervals determined by the critical points. Test points: x = -1, x = 0.5, x = 1.5, and x = 2.Plug in Test Points: Plug test points into g'(x) to check if the derivative is positive (increasing) or negative (decreasing). For x = -1: g'(-1) > 0 (increasing) For x = 0.5: g'(0.5) 0 (increasing) For x = 2: g'(2) 0. Increasing intervals: (-∞, -0.535) and (1, 1.869).

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The derivative of a function
g is given by

g^(')(x)=x^(3)-2x^(2)-3x+4.
On which intervals is the graph of
g increasing?
Use a graphing calculator.
Choose 1 answer:
(A)
x <= -1.562 and

1 <= x <= 2.562
(B)
-1.562 <= x <= 1 and
x >= 2.562
(C)
x <= -0.535 and
x >= 1.869
(D)
-0.535 <= x <= 1.869
(E) All real numbers

The derivative of a function $g$ is given by $\newline$ $g^{\prime}(x)=x^{3}-2 x^{2}-3 x+4 .$ $\newline$ On which intervals is the graph of $g$ increasing? $\newline$ Use a graphing calculator. $\newline$ Choose $1$ answer: $\newline$ (A) $x \leq-1.562$ and $\newline$ $1 \leq x \leq 2.562$ $\newline$ (B) $-1.562 \leq x \leq 1$ and $x \geq 2.562$ $\newline$ (C) $x \leq-0.535$ and $x \geq 1.869$ $\newline$ (D) $-0.535 \leq x \leq 1.869$ $\newline$ (E) All real numbers

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Algebra 2

Ratio and Quadratic equation

Full solution

Q. The derivative of a function

g

is given by

\newline

g^{\prime}(x)=x^{3}-2 x^{2}-3 x+4 .

\newline

On which intervals is the graph of

g

increasing?

\newline

Use a graphing calculator.

\newline

Choose

1

answer:

\newline

(A)

x \leq-1.562

and

\newline

1 \leq x \leq 2.562

\newline

(B)

-1.562 \leq x \leq 1

and

x \geq 2.562

\newline

(C)

x \leq-0.535

and

x \geq 1.869

\newline

(D)

-0.535 \leq x \leq 1.869

\newline

(E) All real numbers

Find Critical Points: Find the critical points of $g'(x)$ by setting the derivative equal to zero and solving for $x$ . $g'(x) = x^3 - 2x^2 - 3x + 4 = 0$
Use Graphing Calculator: Use a graphing calculator to find the roots of the equation. $\newline$ Roots are approximately $x = -0.535$ , $x = 1$ , and $x = 1.869$ .
Determine Sign of $g'(x)$ : Determine the sign of $g'(x)$ on the intervals determined by the critical points. $\newline$ Test points: $x = -1$ , $x = 0.5$ , $x = 1.5$ , and $x = 2$ .
Plug in Test Points: Plug test points into $g'(x)$ to check if the derivative is positive (increasing) or negative (decreasing). $\newline$ For $x = -1$ : g'(-1) > 0 (increasing) $\newline$ For $x = 0.5$ : g'(0.5) < 0 (decreasing) $\newline$ For $x = 1.5$ : g'(1.5) > 0 (increasing) $\newline$ For $x = 2$ : g'(2) < 0 (decreasing)
Graph of g is Increasing: Based on the sign of $g'(x)$ , the graph of $g$ is increasing on the intervals where g'(x) > 0. $\newline$ Increasing intervals: $(-\infty, -0.535)$ and $(1, 1.869)$ .