Question

Differentiate the function. Then find an equation of the tangent line at the indicated point on the graph of the function.

Answer 1

The derivative of the function:

`f'(x)=-(1)/(2√(5-x) )`

The tangent line of the function at the given point:

`y=-(1)/(6)x-(25)/(3)`

Explanation:

Find the equation of the tangent line of the given function using the given point.

`f(x)=6+√(5-x); \ (x,y)=(-4,9)`

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To find the equation of the tangent line to a function at a given point, follow these step-by-step instructions:

Step 1: Identify the point of tangency

• Determine the x-coordinate of the point of tangency. Let's call it x₀.

• Find the corresponding y-coordinate of the point of tangency. Let's call it y₀.

Step 2: Find the derivative of the function

• Calculate the derivative of the given function. Let's denote it as f'(x).

Step 3: Substitute the x-coordinate into the derivative

• Replace the variable x in the derivative function f'(x) with the x-coordinate of the point of tangency (x₀).

• Evaluate the derivative at x₀ to find the slope of the tangent line. Let's denote it as m.

Step 4: Write the equation of the tangent line

• Use the point-slope form of a line: y - y₀ = m(x - x₀).

• Substitute the values of m, x₀, and y₀ into the equation.

• Simplify and rearrange the equation to obtain the final form.

Step 5: Optional - Simplify the equation

• If necessary, simplify the equation by performing any algebraic manipulations.

Step 6: Optional - Verify the equation

• Check the obtained equation by plugging in other points along the tangent line and ensuring they satisfy the equation.

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Step 1:

`(x_0,y_0) \rightarrow (-4,9)`

Step 2:

`f(x)=6+√(5-x)\\\\\\\Longrightarrow f(x)=6+(5-x)^(1/2)\\\\\\\Longrightarrow f'(x)=(1)/(2) (5-x)^(1/2-1) \cdot -1\\\\\\\therefore \boxed{f'(x)=-(1)/(2√(5-x) ) }`

Step 3:

`f'(x)=-(1)/(2√(5-x) ) ; \ (-4,9)\\\\\\\Longrightarrow f'(-4)=-(1)/(2√(5-(-4)) ) \\\\\\\Longrightarrow f'(-4)=-(1)/(2√(9)) \\\\\\\Longrightarrow f'(-4)=-(1)/(2(3)) \\\\\\\therefore \boxed{ f'(-4)=m=-(1)/(6) }`

Step 4 and 5:

`y-y_0=m(x-x_0)\\\\\\\Longrightarrow y-9=-(1)/(6)(x-(-4)) \\\\\\\Longrightarrow y-9=-(1)/(6)(x+4) \\\\\\\Longrightarrow y-9=-(1)/(6)x-(2)/(3)\\\\ \\\therefore \boxed{\boxed{ y=-(1)/(6)x-(25)/(3)}}`

Thus, the problem is solved.