Let f be a function such that f(-1)=3 and f^(')(-1)=5. Let g be the function g(x)=2x^(3). Let F be a function defined as F(x)=(f(x))/(g(x)). F^(')(-1)=

Apply Quotient Rule: To find F'(-1), we need to use the quotient rule for differentiation, which states that if F(x) = f(x) / g(x), then F'(x) = (f'(x)g(x) - f(x)g'(x)) / (g(x))^2. We are given f(-1) = 3 and f'(-1) = 5. We also have g(x) = 2x^3, so we need to find g(-1) and g'(-1).Find g(-1): First, let's find g(-1) by substituting -1 into g(x): g(-1) = 2(-1)^3 g(-1) = 2(-1) g(-1) = -2Find g'(-1): Next, we need to find g'(-1). To do this, we first differentiate g(x) with respect to x: g'(x) = d/dx [2x^3] g'(x) = 2 * 3x^(3-1) g'(x) = 6x^2 Now we substitute -1 into g'(x) to find g'(-1): g'(-1) = 6(-1)^2 g'(-1) = 6(1) g'(-1) = 6Apply Quotient Rule: Now we have all the values we need to apply the quotient rule: F'(-1) = (f'(-1)g(-1) - f(-1)g'(-1)) / (g(-1))^2 F'(-1) = (5 * -2 - 3 * 6) / (-2)^2Simplify Numerator and Denominator: Let's simplify the numerator and the denominator separately: Numerator: 5 * -2 - 3 * 6 = -10 - 18 = -28 Denominator: (-2)^2 = 4Find F'(-1): Now we can find F'(-1): F'(-1) = -28 / 4 F'(-1) = -7

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Let
f be a function such that
f(-1)=3 and
f^(')(-1)=5.
Let
g be the function
g(x)=2x^(3).

Let
F be a function defined as
F(x)=(f(x))/(g(x)).

F^(')(-1)=

- Let $f$ be a function such that $f(-1)=3$ and $f^{\prime}(-1)=5$ . $\newline$ - Let $g$ be the function $g(x)=2 x^{3}$ . $\newline$ Let $F$ be a function defined as $F(x)=\frac{f(x)}{g(x)}$ . $\newline$ $F^{\prime}(-1)=$

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Q. - Let

f

be a function such that

f(-1)=3

and

f^{\prime}(-1)=5

\newline

- Let

g

be the function

g(x)=2 x^{3}

\newline

Let

F

be a function defined as

F(x)=\frac{f(x)}{g(x)}

\newline

F^{\prime}(-1)=

Apply Quotient Rule: To find $F'(-1)$ , we need to use the quotient rule for differentiation, which states that if $F(x) = \frac{f(x)}{g(x)}$ , then $F'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ . We are given $f(-1) = 3$ and $f'(-1) = 5$ . We also have $g(x) = 2x^3$ , so we need to find $g(-1)$ and $g'(-1)$ .
Find $g(-1)$ : First, let's find $g(-1)$ by substituting $-1$ into $g(x)$ :
$g(-1) = 2(-1)^3$
$g(-1) = 2(-1)$
$g(-1) = -2$
Find $g'(-1)$ : Next, we need to find $g'(-1)$ . To do this, we first differentiate $g(x)$ with respect to $x$ :
$g'(x) = \frac{d}{dx} [2x^3]$
$g'(x) = 2 \cdot 3x^{3-1}$
$g'(x) = 6x^2$
Now we substitute $-1$ into $g'(x)$ to find $g'(-1)$ :
$g'(-1)$ $0$
$g'(-1)$ $1$
$g'(-1)$ $2$
Apply Quotient Rule: Now we have all the values we need to apply the quotient rule: $\newline$ $F'(-1) = \frac{f'(-1)g(-1) - f(-1)g'(-1)}{(g(-1))^2}$ $\newline$ $F'(-1) = \frac{(5 \cdot -2 - 3 \cdot 6)}{(-2)^2}$
Simplify Numerator and Denominator: Let's simplify the numerator and the denominator separately: $\newline$ Numerator: $5 \times -2 - 3 \times 6 = -10 - 18 = -28$ $\newline$ Denominator: $(-2)^2 = 4$
Find $F'(-1)$ : Now we can find $F'(-1)$ : $F'(-1) = \frac{-28}{4}$ $F'(-1) = -7$