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In Exercises \(45-48\) a. Find the derivative \(f^{\prime}(x)\) of the given function \(y=f(x)\) b. Graph \(y=f(x)\) and \(y=f^{\prime}(x)\) side by side using separate sets coordinate axes, and answer the following questions. c. For what values of \(x,\) if any, is \(f^{\prime}\) positive? Zero? Negative? d. Over what intervals of \(x\) -values, if any, does the function \(y=f(x)\) increase as \(x\) increases? Decrease as \(x\) increases? How is this related to what you found in part (c)? (We will say more about this relationship in Chapter \(4 . )\) $$ y=-x^{2} $$

Short Answer

Expert verified
Derivative is \( f'(x) = -2x \); \( f'(x) \) is zero at \( x = 0 \), positive when \( x < 0 \), and negative when \( x > 0 \).

Step by step solution

01

Find the Derivative

Start with the given function \( y = f(x) = -x^2 \). To find the derivative \( f'(x) \), use the power rule for derivatives, which states: \( \frac{d}{dx} [x^n] = nx^{n-1} \). Apply this to \( -x^2 \):\[ f'(x) = \frac{d}{dx} (-x^2) = -2x^{2-1} = -2x \].Thus, the derivative of the function is \( f'(x) = -2x \).
02

Graph the Functions

Graph \( y = f(x) = -x^2 \), which is a downward-opening parabola with vertex at the origin. Next, graph the derivative \( y = f'(x) = -2x \), which is a straight line with slope \(-2\) passing through the origin. Ensure each function is on separate axes but side by side to compare them directly.
03

Finding Zeroes of f'(x)

To find where \( f'(x) \) is zero, solve \( -2x = 0 \) for \( x \). Divide both sides by \(-2\) to get \( x = 0 \). So, \( f'(x) \) is zero at \( x = 0 \).
04

Identifying Where f'(x) is Positive or Negative

For \( f'(x) = -2x \), determine when the derivative is positive or negative:- \( f'(x) = -2x > 0 \) when \( x < 0 \).- \( f'(x) = -2x < 0 \) when \( x > 0 \).
05

Determine Intervals of Increase and Decrease in f(x)

When \( f'(x) > 0 \), the function \( y = f(x) \) is increasing, and when \( f'(x) < 0 \), it is decreasing:- \( y = f(x) = -x^2 \) increases as \( x \) increases for \( x < 0 \), because \( f'(x) > 0 \) in this interval.- \( y = f(x) = -x^2 \) decreases as \( x \) increases for \( x > 0 \), since \( f'(x) < 0 \) in this interval.This behavior corresponds directly to the sign of \( f'(x) \).

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Key Concepts

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

Power Rule for Derivatives
The power rule is a fundamental tool for finding the derivative of a function. It is particularly useful when dealing with functions of the form \( x^n \). The rule states: if \( y = x^n \), then the derivative \( \frac{d}{dx}[x^n] = nx^{n-1} \).

So, for the function \( y = -x^2 \), applying the power rule means bringing down the exponent 2, multiplying by the coefficient (which is -1 in this case), and reducing the exponent by 1. Hence, we derive:
  • \( f'(x) = \frac{d}{dx} [-x^2] = -2x \).
By understanding the power rule, we can easily find derivatives of polynomial functions. Knowing derivatives helps in analyzing the behavior of functions, such as increasing or decreasing trends, and points of inflection.

This straightforward method is key in calculus, allowing us to deduce complex changes in functions smoothly.
Graphing Functions
Graphing a function and its derivative provides a visual insight into the function's behavior.
  • The function \( y = -x^2 \) is a downward-opening parabola with its vertex at the origin. This shape is characteristic of a quadratic with a negative leading coefficient.
  • On the other hand, its derivative \( f'(x) = -2x \) is a straight line passing through the origin, with a negative slope of -2.
Graphing them side by side helps in directly comparing the original function to how it changes:
  • The slope of the tangent (given by the derivative) tells us how steep the parabola is at any point on its curve.
  • When graphing these, ensure each graph is on separate axes to clearly see how the behavior of the derivative influences the graph of the original function.
This visualization between a function and its derivative aids in understanding the nature of the changes occurring in the function.
Intervals of Increase and Decrease
The intervals of increase and decrease tell us about the behavior of a function as \( x \) changes. If \( f'(x) > 0 \), the function \( y = f(x) \) is increasing. Conversely, if \( f'(x) < 0 \), it is decreasing.

For the function \( y = -x^2 \):
  • When \( x < 0 \), \( f'(x) = -2x > 0 \). This means the function is increasing in the interval from negative infinity up to zero.
  • When \( x > 0 \), \( f'(x) = -2x < 0 \). Thus, the function is decreasing for positive values of \( x \).
By identifying where the derivative equals zero (at \( x = 0 \)), we note that the graph reaches a vertex point at this location, which is often a maximum for parabolas like \( y = -x^2 \).

This relationship is crucial in calculus for understanding where a function reaches its peaks and troughs, and it connects directly to where the derivative changes sign.

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Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ x^{3}-x y+y^{3}=7, \quad P(2,1) $$

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the lateral surface area \(S=\pi r \sqrt{r^{2}+h^{2}}\) of a right circular cone when the radius changes from \(r_{0}\) to \(r_{0}+d r\) and the height does not change

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

The cissoid of Diocles (from about 200 B.C.) Find equations for the tangent and normal to the cissoid of Diocles \(y^{2}(2-x)=x^{3}\) at \((1,1) .\)

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ x^{2} y^{2}=9, \quad(-1,3) $$

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