Question

Derivatives and tangent lines a. For the following functions and values of $a,$ find $f^{\prime }(a)$ b. Determine an equation of the line tangent to the graph of $f$ at the point $(a,f(a))$ for the given value of $a$ $$f(x)=\frac{1}{\sqrt{x}};a=\frac{1}{4}$$

Short answer

Answer: The equation of the tangent line is $y=-4x+3$.

Step-by-step solution

Calculate the derivative of the function

First, we find the derivative $f^{\prime }$ of the given function $f(x)=\frac{1}{\sqrt{x}}$. To do this, we rewrite the function as $f(x)=x^{-1/2}$ and apply the power rule, which states that the derivative of $x^n$ with respect to $x$ is $n{x}^{n-1}$. $$f^{\prime }(x)=\frac{d}{dx}(x^{-1/2})=-\frac{1}{2}{x}^{-3/2}$$

Evaluate the derivative at the given point

Now, we need to evaluate $f^{\prime }(a)$ at the given point $a=\frac{1}{4}$. Substitute $a$ into the derived expression: $$f^{\prime }(\frac{1}{4})=-\frac{1}{2}(\frac{1}{4}{)}^{-3/2}$$ Simplify to obtain the slope $m$: $$m=-\frac{1}{2}(\frac{1}{4}{)}^{-3/2}=-\frac{1}{2}(8)=-4$$

Find the function value at the given point

Now we need to find the value of the original function at the given point $a=\frac{1}{4}$, i.e., we need to find $f(\frac{1}{4})$: $$f(\frac{1}{4})=\frac{1}{\sqrt{\frac{1}{4}}}=\frac{1}{\frac{1}{2}}=2$$ So, the point $(a,f(a))$ on the tangent line is $(\frac{1}{4},2)$.

Determine the equation of the tangent line

Now we have all the necessary information to find the equation of the tangent line. We know the point $(a,f(a))$ on the tangent line is $(\frac{1}{4},2)$ and the slope $m=-4$. We use the point-slope form to find the equation of the tangent line: $$y-f(a)=m(x-a)$$ Substitute the known values: $$y-2=-4(x-\frac{1}{4})$$ To obtain the final equation for the tangent line, expand and simplify: $$y-2=-4x+1$$ $$y=-4x+3$$ So, the equation of the tangent line is $y=-4x+3$.

Key concepts

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function is changing at any given point. Imagine driving a car: your speedometer shows your instantaneous rate of change of distance with respect to time – that's a real-life derivative! In mathematical terms, the derivative of a function measures how the output value of the function changes as its input value changes. The process of finding a derivative is called differentiation.

For a function represented as $y=f(x)$, the notation for its derivative is $f^{\prime }(x)$ or $\frac{dy}{dx}$. When solving problems related to derivatives, you are often looking to find the slope of the function at a particular point, which gives you the steepness or incline of the tangent line at that point.

Power Rule

The power rule is a shortcut in differentiation that makes finding the derivative of polynomial functions much easier. It applies when you have a function of the form $f(x)=x^n$, where $n$ is any real number. The power rule states that the derivative of this function is $n{x}^{n-1}$.

For instance, applying the power rule to the function $f(x)=x^3$, you get $f^{\prime }(x)=3x^{3-1}=3x^2$. It's essential to remember this rule as it simplifies the process of differentiation, especially when dealing with higher powers of $x$. In the given exercise, the power rule was used to determine the derivative of $f(x)=x^{-1/2}$, leading to a derivative of $f^{\prime }(x)=-\frac{1}{2}{x}^{-3/2}$, demonstrating its versatility and ease of use.

Point-Slope Form

The point-slope form is one of the forms used to express the equation of a straight line. It is particularly useful when you know a point on the line $(x_1,y_1)$ and the line’s slope $m$. The formula for the point-slope form is given by:$$y-y_1=m(x-x_1)$$.

You can think of this formula as a way to trace every point on the line by starting from the known point $(x_1,y_1)$ and moving $m$ units in the y-direction for each unit moved in the x-direction. This is exactly what we did in the exercise when determining the equation of the tangent line by plugging in the slope $m=-4$ and the point $(\frac{1}{4},2)$.

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input. For instance, if you have a function $f(x)$, evaluating it at $x=a$ is done by replacing every occurrence of $x$ in the function with $a$. The result, notated as $f(a)$, is the value of the function at that particular point.

Using the original exercise as an example, we evaluated the function $f(x)=\frac{1}{\text{/}x}$ at $x=\frac{1}{4}$ to find $f(\frac{1}{4})=2$. This step is vital, as knowing the function value at the point where you’re applying the tangent line informs the exact position of the line in relation to the graph of the function.