If the derivative of the function $f(x) =\frac{m}{x} +2nx + 1$ vanishes at x = 2, then what is the value of m + 8n ?

Question

Prepp Team · Accepted Answer

Understanding the Problem: Function, Derivative, and Vanishing Point

The question asks us to consider a function $f(x) =\frac{m}{x} +2nx + 1$. We are given a crucial piece of information: the derivative of this function vanishes (becomes zero) at a specific point, $x = 2$. Our goal is to find the value of the expression $m + 8n$.

To solve this, we first need to find the derivative of $f(x)$ with respect to $x$. Then, we will use the condition that this derivative is zero at $x=2$ to form an equation involving $m$ and $n$. Finally, we will see if this equation allows us to find the value of $m + 8n$.

Calculating the Derivative of f(x)

The function is given by $f(x) = \frac{m}{x} + 2nx + 1$. We can rewrite $\frac{m}{x}$ as $mx^{-1}$. So, $f(x) = mx^{-1} + 2nx + 1$.

Now, let's find the derivative $f'(x)$ using standard differentiation rules:

The derivative of $cx^k$ is $ckx^{k-1}$.
The derivative of a constant is $0$.

Applying these rules: \[ f'(x) = \frac{d}{dx}(mx^{-1}) + \frac{d}{dx}(2nx) + \frac{d}{dx}(1) \] \[ f'(x) = m(-1)x^{-1-1} + 2n(1)x^{1-1} + 0 \] \[ f'(x) = -mx^{-2} + 2nx^0 \] Since $x^0 = 1$ (for $x \neq 0$), we get: \[ f'(x) = -\frac{m}{x^2} + 2n \] This is the derivative of the function $f(x)$.

Using the Vanishing Condition at x = 2

We are told that the derivative vanishes at $x = 2$. This means $f'(2) = 0$. Let's substitute $x = 2$ into our expression for $f'(x)$: \[ f'(2) = -\frac{m}{(2)^2} + 2n \] \[ 0 = -\frac{m}{4} + 2n \]

Now we have an equation relating $m$ and $n$: \[ -\frac{m}{4} + 2n = 0 \]

Analyzing the Equation and Finding m + 8n

Let's rearrange the equation $-\frac{m}{4} + 2n = 0$ to see the relationship between $m$ and $n$: \[ 2n = \frac{m}{4} \] To eliminate the fraction, we can multiply both sides by 4: \[ 4 \times (2n) = 4 \times \left(\frac{m}{4}\right) \] \[ 8n = m \] So, we found that $m$ is equal to $8n$.

The question asks for the value of $m + 8n$. Let's use the relationship $m = 8n$ in this expression: \[ m + 8n \] Substitute $m$ with $8n$: \[ (8n) + 8n = 16n \] Alternatively, substitute $8n$ with $m$: \[ m + (m) = 2m \]

The value of the expression $m + 8n$ is $16n$ (or $2m$). However, we have only one equation ($m=8n$) relating the two variables $m$ and $n$. We cannot determine the specific numerical values of $m$ or $n$ individually. Since the value of $m + 8n$ depends on the value of $n$ (or $m$), it can take different values depending on which pair of $m$ and $n$ satisfying $m=8n$ is chosen.

For instance:

If $n=1$, then $m=8(1)=8$. $m+8n = 8 + 8(1) = 16$.
If $n=2$, then $m=8(2)=16$. $m+8n = 16 + 8(2) = 32$.
If $n=0$, then $m=8(0)=0$. $m+8n = 0 + 8(0) = 0$.

Since the value of $m + 8n$ is not a unique number, it cannot be determined from the information given.

Conclusion on Determining m + 8n

Based on our analysis, the condition that the derivative vanishes at $x=2$ gives us the relationship $m = 8n$. However, this single equation is not enough to find the individual values of $m$ and $n$, and consequently, it is not enough to find a unique numerical value for the expression $m + 8n$. The value of $m + 8n$ is dependent on the values of $m$ or $n$ that satisfy the relationship.

Therefore, the value of $m + 8n$ cannot be determined due to insufficient data.

Step	Action	Result
1	Find the derivative $f'(x)$	$f'(x) = -\frac{m}{x^2} + 2n$
2	Set $f'(2) = 0$	$-\frac{m}{4} + 2n = 0$
3	Solve for relationship between $m$ and $n$	$m = 8n$
4	Evaluate $m + 8n$	$m + 8n = (8n) + 8n = 16n$ (or $2m$)
5	Check if value is unique	Value depends on $n$ (or $m$); not unique.

Revision Table: Key Calculus Concepts

Concept	Definition/Explanation	Application in Problem
Derivative	Measures the instantaneous rate of change of a function. Represents the slope of the tangent line at a point. Notation: $f'(x)$ or $\frac{dy}{dx}$.	Calculated $f'(x)$ to use the given condition.
Power Rule of Differentiation	If $f(x) = x^k$, then $f'(x) = kx^{k-1}$. Applied to terms like $x^{-1}$ and $x^1$.	Used to differentiate $mx^{-1}$ and $2nx$.
Derivative of a Constant	If $f(x) = c$ (where c is a constant), then $f'(x) = 0$.	Used for the constant term $1$ in $f(x)$.
Vanishing Derivative	When $f'(a) = 0$ for some point $x=a$, it indicates a critical point. Critical points can correspond to local maximums, minimums, or inflection points with a horizontal tangent.	Used the condition $f'(2) = 0$ to create an equation.

Additional Information: Critical Points and Optimization

The point where the derivative of a function vanishes ($f'(x)=0$) is called a critical point. Critical points are very important in calculus, especially in optimization problems, where we try to find the maximum or minimum values of a function.

At a critical point, the tangent line to the graph of the function is horizontal. While a vanishing derivative indicates a critical point, it doesn't automatically mean it's a local maximum or minimum. Further tests, like the second derivative test or analyzing the sign change of the first derivative, are needed to classify a critical point.

In this specific problem, we used the critical point information ($f'(2)=0$) to establish a relationship between the parameters $m$ and $n$. However, without more information (like another point the function passes through, or another condition on the derivative), we couldn't find unique values for $m$ and $n$, which is why the expression $m + 8n$ remained undetermined as a single numerical value.

Prepp Team · Answer

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Prepp Team · Answer

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Prepp Team · Answer

2

Step	Action	Result
1	Find the derivative \(f'(x)\)	\(f'(x) = -\frac{m}{x^2} + 2n\)
2	Set \(f'(2) = 0\)	\(-\frac{m}{4} + 2n = 0\)
3	Solve for relationship between \(m\) and \(n\)	\(m = 8n\)
4	Evaluate \(m + 8n\)	\(m + 8n = (8n) + 8n = 16n\) (or \(2m\))
5	Check if value is unique	Value depends on \(n\) (or \(m\)); not unique.

Concept	Definition/Explanation	Application in Problem
Derivative	Measures the instantaneous rate of change of a function. Represents the slope of the tangent line at a point. Notation: \(f'(x)\) or \(\frac{dy}{dx}\).	Calculated \(f'(x)\) to use the given condition.
Power Rule of Differentiation	If \(f(x) = x^k\), then \(f'(x) = kx^{k-1}\). Applied to terms like \(x^{-1}\) and \(x^1\).	Used to differentiate \(mx^{-1}\) and \(2nx\).
Derivative of a Constant	If \(f(x) = c\) (where c is a constant), then \(f'(x) = 0\).	Used for the constant term \(1\) in \(f(x)\).
Vanishing Derivative	When \(f'(a) = 0\) for some point \(x=a\), it indicates a critical point. Critical points can correspond to local maximums, minimums, or inflection points with a horizontal tangent.	Used the condition \(f'(2) = 0\) to create an equation.