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

Given values: We are given g(1) = 5 and g'(1) = -3. We also know that h(x) = x^2, so h(1) = 1^2 = 1 and h'(x) = 2x, so h'(1) = 2*1 = 2.Quotient rule for derivatives: To find F'(x), we need to use the quotient rule for derivatives, which states that if F(x) = g(x)/h(x), then F'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^2.Substitute values into formula: Now we substitute x = 1 into the quotient rule formula using the values we have: g(1) = 5, g'(1) = -3, h(1) = 1, and h'(1) = 2. F'(1) = ((-3)*1 - 5*2)/(1)^2 = (-3 - 10)/1 = -13/1 = -13.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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

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

F^(')(1)=

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

View step-by-step help

Home

Math Problems

Algebra 2

Find the vertex of the transformed function

Full solution

Q. - Let

g

be a function such that

g(1)=5

and

g^{\prime}(1)=-3

\newline

- Let

h

be the function

h(x)=x^{2}

\newline

Let

F

be a function defined as

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

\newline

F^{\prime}(1)=

Given values: We are given $g(1) = 5$ and $g'(1) = -3$ . We also know that $h(x) = x^2$ , so $h(1) = 1^2 = 1$ and $h'(x) = 2x$ , so $h'(1) = 2\cdot 1 = 2$ .
Quotient rule for derivatives: To find $F'(x)$ , we need to use the quotient rule for derivatives, which states that if $F(x) = \frac{g(x)}{h(x)}$ , then $F'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$ .
Substitute values into formula: Now we substitute $x = 1$ into the quotient rule formula using the values we have: $g(1) = 5$ , $g'(1) = -3$ , $h(1) = 1$ , and $h'(1) = 2$ . $F'(1) = ((-3)\cdot 1 - 5\cdot 2)/(1)^2 = (-3 - 10)/1 = -13/1 = -13$ .