Solved on Feb 24, 2024

Calculate $g(4)$ and $g'(4)$ where $g(x)$ is the inverse of $f(x)=\sqrt{x^2+6x}$ for $x \geq 0$ .

STEP 1

Assumptions
1. The function $f(x) = \sqrt{x^2 + 6x}$ is defined for $x \geq 0$ .
2. The function $g(x)$ is the inverse of $f(x)$ .
3. We need to find $g(4)$ and $g'(4)$ , where $g'(x)$ is the derivative of $g(x)$ .

STEP 2

To find $g(4)$ , we need to solve $f(x) = 4$ for $x$ since $g(f(x)) = x$ for the inverse function.
$f(x) = \sqrt{x^2 + 6x} = 4$

STEP 3

Square both sides of the equation to eliminate the square root.
$(\sqrt{x^2 + 6x})^2 = 4^2$
$x^2 + 6x = 16$

STEP 4

Rearrange the equation to standard quadratic form.
$x^2 + 6x - 16 = 0$

STEP 5

Factor the quadratic equation to find the values of $x$ .
$(x + 8)(x - 2) = 0$

STEP 6

Solve for $x$ .
$x + 8 = 0 \quad \text{or} \quad x - 2 = 0$
$x = -8 \quad \text{or} \quad x = 2$

STEP 7

Since the domain of $f(x)$ is $x \geq 0$ , we discard the negative solution.
$x = 2$

STEP 8

Thus, $g(4) = 2$ .

STEP 9

To find $g'(4)$ , we use the formula for the derivative of an inverse function at a point:
$g'(a) = \frac{1}{f'(g(a))}$

STEP 10

First, we find $f'(x)$ , the derivative of $f(x)$ .
$f'(x) = \frac{d}{dx} \sqrt{x^2 + 6x}$

STEP 11

Use the chain rule to differentiate $\sqrt{x^2 + 6x}$ .
$f'(x) = \frac{1}{2\sqrt{x^2 + 6x}} \cdot \frac{d}{dx}(x^2 + 6x)$

STEP 12

Differentiate the inside function $x^2 + 6x$ .
$\frac{d}{dx}(x^2 + 6x) = 2x + 6$

STEP 13

Substitute the derivative of the inside function into the derivative of $f(x)$ .
$f'(x) = \frac{1}{2\sqrt{x^2 + 6x}} \cdot (2x + 6)$

STEP 14

Simplify the expression for $f'(x)$ .
$f'(x) = \frac{2x + 6}{2\sqrt{x^2 + 6x}}$

STEP 15

Now, we need to evaluate $f'(x)$ at $x = g(4) = 2$ .
$f'(2) = \frac{2(2) + 6}{2\sqrt{2^2 + 6(2)}}$

STEP 16

Calculate the values in the numerator and the denominator.
$f'(2) = \frac{4 + 6}{2\sqrt{4 + 12}}$

STEP 17

Simplify the expression.
$f'(2) = \frac{10}{2\sqrt{16}}$

STEP 18

Further simplify the expression.
$f'(2) = \frac{10}{2 \cdot 4}$

STEP 19

Complete the simplification.
$f'(2) = \frac{10}{8} = \frac{5}{4}$

STEP 20

Now, use the formula for the derivative of the inverse function to find $g'(4)$ .
$g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(2)}$

STEP 21

Substitute the value of $f'(2)$ .
$g'(4) = \frac{1}{\frac{5}{4}}$

STEP 22

Calculate $g'(4)$ .
$g'(4) = \frac{4}{5}$

STEP 23

We have found both $g(4)$ and $g'(4)$ .
$\begin{array}{l} g(4) = 2 \\ g^{\prime}(4) = \frac{4}{5} \end{array}$

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Calculate g(4)g(4)g(4) and g′(4)g'(4)g′(4) where g(x)g(x)g(x) is the inverse of f(x)=x2+6xf(x)=\sqrt{x^2+6x}f(x)=x2+6x​ for x≥0x \geq 0x≥0.

STEP 1

Assumptions1. The function f(x)=x2+6xf(x) = \sqrt{x^2 + 6x}f(x)=x2+6x​ is defined for x≥0x \geq 0x≥0.2. The function g(x)g(x)g(x) is the inverse of f(x)f(x)f(x).3. We need to find g(4)g(4)g(4) and g′(4)g'(4)g′(4), where g′(x)g'(x)g′(x) is the derivative of g(x)g(x)g(x).

STEP 2

To find g(4)g(4)g(4), we need to solve f(x)=4f(x) = 4f(x)=4 for xxx since g(f(x))=xg(f(x)) = xg(f(x))=x for the inverse function.f(x)=x2+6x=4f(x) = \sqrt{x^2 + 6x} = 4f(x)=x2+6x​=4

STEP 3

Square both sides of the equation to eliminate the square root.(x2+6x)2=42(\sqrt{x^2 + 6x})^2 = 4^2(x2+6x​)2=42x2+6x=16x^2 + 6x = 16x2+6x=16

STEP 4

Rearrange the equation to standard quadratic form.x2+6x−16=0x^2 + 6x - 16 = 0x2+6x−16=0

STEP 5

Factor the quadratic equation to find the values of xxx.(x+8)(x−2)=0(x + 8)(x - 2) = 0(x+8)(x−2)=0

STEP 6

Solve for xxx.x+8=0orx−2=0x + 8 = 0 \quad \text{or} \quad x - 2 = 0x+8=0orx−2=0x=−8orx=2x = -8 \quad \text{or} \quad x = 2x=−8orx=2

STEP 7

Since the domain of f(x)f(x)f(x) is x≥0x \geq 0x≥0, we discard the negative solution.x=2x = 2x=2

STEP 8

Thus, g(4)=2g(4) = 2g(4)=2.

STEP 9

To find g′(4)g'(4)g′(4), we use the formula for the derivative of an inverse function at a point:g′(a)=1f′(g(a))g'(a) = \frac{1}{f'(g(a))}g′(a)=f′(g(a))1​

STEP 10

First, we find f′(x)f'(x)f′(x), the derivative of f(x)f(x)f(x).f′(x)=ddxx2+6xf'(x) = \frac{d}{dx} \sqrt{x^2 + 6x}f′(x)=dxd​x2+6x​

STEP 11

Use the chain rule to differentiate x2+6x\sqrt{x^2 + 6x}x2+6x​.f′(x)=12x2+6x⋅ddx(x2+6x)f'(x) = \frac{1}{2\sqrt{x^2 + 6x}} \cdot \frac{d}{dx}(x^2 + 6x)f′(x)=2x2+6x​1​⋅dxd​(x2+6x)

STEP 12

Differentiate the inside function x2+6xx^2 + 6xx2+6x.ddx(x2+6x)=2x+6\frac{d}{dx}(x^2 + 6x) = 2x + 6dxd​(x2+6x)=2x+6

STEP 13

Substitute the derivative of the inside function into the derivative of f(x)f(x)f(x).f′(x)=12x2+6x⋅(2x+6)f'(x) = \frac{1}{2\sqrt{x^2 + 6x}} \cdot (2x + 6)f′(x)=2x2+6x​1​⋅(2x+6)

STEP 14

Simplify the expression for f′(x)f'(x)f′(x).f′(x)=2x+62x2+6xf'(x) = \frac{2x + 6}{2\sqrt{x^2 + 6x}}f′(x)=2x2+6x​2x+6​

STEP 15

Now, we need to evaluate f′(x)f'(x)f′(x) at x=g(4)=2x = g(4) = 2x=g(4)=2.f′(2)=2(2)+6222+6(2)f'(2) = \frac{2(2) + 6}{2\sqrt{2^2 + 6(2)}}f′(2)=222+6(2)​2(2)+6​

STEP 16

Calculate the values in the numerator and the denominator.f′(2)=4+624+12f'(2) = \frac{4 + 6}{2\sqrt{4 + 12}}f′(2)=24+12​4+6​

STEP 17

Simplify the expression.f′(2)=10216f'(2) = \frac{10}{2\sqrt{16}}f′(2)=216​10​

STEP 18

Further simplify the expression.f′(2)=102⋅4f'(2) = \frac{10}{2 \cdot 4}f′(2)=2⋅410​

STEP 19

Complete the simplification.f′(2)=108=54f'(2) = \frac{10}{8} = \frac{5}{4}f′(2)=810​=45​

STEP 20

Now, use the formula for the derivative of the inverse function to find g′(4)g'(4)g′(4).g′(4)=1f′(g(4))=1f′(2)g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(2)}g′(4)=f′(g(4))1​=f′(2)1​

STEP 21

Substitute the value of f′(2)f'(2)f′(2).g′(4)=154g'(4) = \frac{1}{\frac{5}{4}}g′(4)=45​1​

STEP 22

Calculate g′(4)g'(4)g′(4).g′(4)=45g'(4) = \frac{4}{5}g′(4)=54​

STEP 23

We have found both g(4)g(4)g(4) and g′(4)g'(4)g′(4).g(4)=2g′(4)=45 \begin{array}{l} g(4) = 2 \\ g^{\prime}(4) = \frac{4}{5} \end{array} g(4)=2g′(4)=54​​

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Calculate $g(4)$ and $g'(4)$ where $g(x)$ is the inverse of $f(x)=\sqrt{x^2+6x}$ for $x \geq 0$ .

Assumptions
1. The function $f(x) = \sqrt{x^2 + 6x}$ is defined for $x \geq 0$ .
2. The function $g(x)$ is the inverse of $f(x)$ .
3. We need to find $g(4)$ and $g'(4)$ , where $g'(x)$ is the derivative of $g(x)$ .

To find $g(4)$ , we need to solve $f(x) = 4$ for $x$ since $g(f(x)) = x$ for the inverse function.
$f(x) = \sqrt{x^2 + 6x} = 4$

Square both sides of the equation to eliminate the square root.
$(\sqrt{x^2 + 6x})^2 = 4^2$
$x^2 + 6x = 16$

Rearrange the equation to standard quadratic form.
$x^2 + 6x - 16 = 0$

Factor the quadratic equation to find the values of $x$ .
$(x + 8)(x - 2) = 0$

Solve for $x$ .
$x + 8 = 0 \quad \text{or} \quad x - 2 = 0$
$x = -8 \quad \text{or} \quad x = 2$

Since the domain of $f(x)$ is $x \geq 0$ , we discard the negative solution.
$x = 2$

Thus, $g(4) = 2$ .

To find $g'(4)$ , we use the formula for the derivative of an inverse function at a point:
$g'(a) = \frac{1}{f'(g(a))}$

First, we find $f'(x)$ , the derivative of $f(x)$ .
$f'(x) = \frac{d}{dx} \sqrt{x^2 + 6x}$

Use the chain rule to differentiate $\sqrt{x^2 + 6x}$ .
$f'(x) = \frac{1}{2\sqrt{x^2 + 6x}} \cdot \frac{d}{dx}(x^2 + 6x)$

Differentiate the inside function $x^2 + 6x$ .
$\frac{d}{dx}(x^2 + 6x) = 2x + 6$

Substitute the derivative of the inside function into the derivative of $f(x)$ .
$f'(x) = \frac{1}{2\sqrt{x^2 + 6x}} \cdot (2x + 6)$

Simplify the expression for $f'(x)$ .
$f'(x) = \frac{2x + 6}{2\sqrt{x^2 + 6x}}$

Now, we need to evaluate $f'(x)$ at $x = g(4) = 2$ .
$f'(2) = \frac{2(2) + 6}{2\sqrt{2^2 + 6(2)}}$

Calculate the values in the numerator and the denominator.
$f'(2) = \frac{4 + 6}{2\sqrt{4 + 12}}$

Simplify the expression.
$f'(2) = \frac{10}{2\sqrt{16}}$

Further simplify the expression.
$f'(2) = \frac{10}{2 \cdot 4}$

Complete the simplification.
$f'(2) = \frac{10}{8} = \frac{5}{4}$

Now, use the formula for the derivative of the inverse function to find $g'(4)$ .
$g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(2)}$

Substitute the value of $f'(2)$ .
$g'(4) = \frac{1}{\frac{5}{4}}$

Calculate $g'(4)$ .
$g'(4) = \frac{4}{5}$

We have found both $g(4)$ and $g'(4)$ .
$\begin{array}{l} g(4) = 2 \\ g^{\prime}(4) = \frac{4}{5} \end{array}$