Question

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

Answer 1

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:

$\rightarrow \sf \dfrac{dy}{dx} = \dfrac{d}{dx} ( \sqrt{12-x} )$

$\rightarrow \sf \dfrac{dy}{dx} = \dfrac{d}{dx} ( (12-x)^{\frac{1}{2} }} )$

$\rightarrow \sf \dfrac{dy}{dx} = \dfrac{1}{2} ( (12-x)^{\frac{1}{2}-1 }} )(-1)$

$\rightarrow \sf \dfrac{1}{2\sqrt{12-x}}\left(-1\right)$

$\rightarrow \sf -\dfrac{1}{2\sqrt{12-x}}$

$\bold{\star}$ Tangent/Parallel Line Has Same Slope

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

$\rightarrow \sf -\dfrac{1}{2\sqrt{12-(-52)}} \ = \ -\dfrac{1}{16} \ \ = \ -0.0625$

Answer 2

Answer:

$-\dfrac{1}{16}$  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.

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

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

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