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Find the slope of the tangent line to f(x) = √(12-x) at x = -52

Write your answer as an integer or a fraction

Please give a thorough answer, this is practice for finals and I have forgotten how to do this.

Respuesta :

fieryanswererft

Answer:

slope: -0.0625

Explanation:

To find slope of a function, we have to find the derivative of the function.

  • Here given the function:  f(x) = √(12-x)

Start deriving the equation:

[tex]\rightarrow \sf \dfrac{dy}{dx} = \dfrac{d}{dx} ( \sqrt{12-x} )[/tex]

[tex]\rightarrow \sf \dfrac{dy}{dx} = \dfrac{d}{dx} ( (12-x)^{\frac{1}{2} }} )[/tex]

[tex]\rightarrow \sf \dfrac{dy}{dx} = \dfrac{1}{2} ( (12-x)^{\frac{1}{2}-1 }} )(-1)[/tex]

[tex]\rightarrow \sf \dfrac{1}{2\sqrt{12-x}}\left(-1\right)[/tex]

[tex]\rightarrow \sf -\dfrac{1}{2\sqrt{12-x}}[/tex]

[tex]\bold{\star}[/tex] Tangent/Parallel Line Has Same Slope

To find slope at x = -52, simplify insert x = -52 in derivative we found

[tex]\rightarrow \sf -\dfrac{1}{2\sqrt{12-(-52)}} \ = \ -\dfrac{1}{16} \ \ = \ -0.0625[/tex]

semsee45

Answer:

[tex]-\dfrac{1}{16}[/tex]  or -0.0625

Step-by-step explanation:

To find the slope of the tangent line at a point, differentiate the function (using the chain rule for this particular function), then input the x-coordinate of the point into the first derivative.

[tex]\begin{aligned}f(x) & = \sqrt{12-x}\\& = (12-x)^{\frac{1}{2}}\\\\\implies f'(x) & =\dfrac{1}{2}(12-x)^{-\frac{1}{2}} \cdot -1\\& = -\dfrac{1}{2\sqrt{12-x}}\end{aligned}[/tex]

Therefore, the slope of the tangent line to f(x) at x = -52 is:

[tex]\begin{aligned}f'(-52) &=-\dfrac{1}{2\sqrt{12-(-52)}}\\\\&=-\dfrac{1}{2\sqrt{64}}\\\\&=-\dfrac{1}{2 \cdot 8}\\\\&=-\dfrac{1}{16}\\\\&=-0.0625\end{aligned}[/tex]

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