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Chapter 3: Problem 35

Find the point(s) on the graph of \(y=\left(x^{2}+3 x-1\right) / x\) where the slope is \(5.\)

Short Answer

Expert verified
There are no points on the graph where the slope is 5.

Step by step solution

01

- Finding the derivative

To find where the slope is 5, calculate the derivative of the function. The function is given as \(y = \frac{x^2 + 3x - 1}{x}\). Simplify it first: \(y = x + 3 - \frac{1}{x}\). Now, take the derivative of \(y\) with respect to \(x\): \(y' = 1 - \frac{1}{x^2}\).
02

- Setting the derivative equal to 5

Set the derivative equal to the given slope (5): \(1 - \frac{1}{x^2} = 5\).
03

- Solving for \(x\)

Solve the equation for \(x\): \[-\frac{1}{x^2} = 4\] \[\frac{1}{x^2} = -4\]. Since \(\frac{1}{x^2}\) cannot be negative, there are no real solutions to this equation.
04

- Conclusion

Since there are no real solutions for \(x\), there are no points on the graph of the function where the slope is 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

finding derivative
To determine where the slope of a function equals 5, we first need to calculate the derivative. The given function is: \[ y = \frac{x^2 + 3x - 1}{x} \]

Before taking the derivative, simplify the function to make it easier to differentiate. Simplified, it becomes: \[ y = x + 3 - \frac{1}{x} \]

Now, differentiate each term separately:
* The derivative of \(x\) is 1.
* The derivative of \(3\) is 0.
* The derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\).
After differentiation, we get: \[ y' = 1 - \frac{1}{x^2} \] This is the expression for the derivative of the function.
solving equations
Next, set the derived function equal to 5, because we want to find the point where the slope is 5.
The equation now is: \[ 1 - \frac{1}{x^2} = 5 \]
To solve for \(x\), first isolate the term involving \(x\): \[ -\frac{1}{x^2} = 4 \] This simplifies to: \[ \frac{1}{x^2} = -4 \] However, we encounter an issue here because \( \frac{1}{x^2} \) must always be positive or zero, and can never equal a negative value like -4. Therefore, this equation has no real solution.
slope of a function
The slope of a function at any point is given by its derivative. For the given function: \[ y = \frac{x^2 + 3x - 1}{x} \]
We found the derivative to be: \[ y' = 1 - \frac{1}{x^2} \]
The slope of the function at any given \(x\) value can be determined by substituting \(x\) into this derivative.
The slope tells us how steep the graph is at that specific point. For a polynomial function, different values of \(x\) will yield different slopes depending on the curvature of the graph at that point.
graph analysis
Graph analysis involves examining the function鈥檚 behavior visually.
By plotting \( y = \frac{x^2 + 3x - 1}{x} \), you'll notice how the graph behaves at different intervals.
Important aspects to consider in graph analysis include:
* Where the graph increases or decreases
* Points of inflection
* Asymptotic behavior, if any
For this graph, because the derivative cannot equal 5, we observe there are no points where the slope is exactly 5.
The graph analysis thus reinforces what we found algebraically. Understanding how the graph looks provides additional insights into the function鈥檚 characteristics.

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Most popular questions from this chapter

Use implicit differentiation of the equations to determine the slope of the graph at the given point. $$4 y^{3}-x^{2}=-5 ; x=3, y=1$$

Compute \(\frac{d}{d x} f(g(x)),\) where \(f(x)\) and \(g(x)\) are the following: $$f(x)=\sqrt{x}, g(x)=x^{2}+1$$

A manufacturer of microcomputers estimates that \(t\) months from now it will sell \(x\) thousand units of its main line of microcomputers per month, where \(x=.05 t^{2}+2 t+5.\) Because of economies of scale, the profit \(P\) from manufacturing and selling \(x\) thousand units is estimated to be \(P=.001 x^{2}+.1 x-.25\) million dollars. Calculate the rate at which the profit will be increasing 5 months from now.

Let \(f(x)=1 / x\) and \(g(x)=x^{3}\). (a) Show that the product rule yields the correct derivative of \((1 / x) x^{3}=x^{2}\). (b) Compute the product \(f^{\prime}(x) g^{\prime}(x),\) and note that it is not the derivative of \(f(x) g(x)\).

In an expression of the form \(f(g(x)), f(x)\) is called the outer function and \(g(x)\) is called the inner function. Give a written description of the chain rule using the words inner and outer.
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