91Ó°ÊÓ

For a certain function \(f\), its derivative is known to be \(f^{\prime}(x)=(x-1) e^{-x^{2}}\). Note that you do not know a formula for \(y=f(x)\). a. At what \(x\) -value(s) is \(f^{\prime}(x)=0\) ? Justify your answer algebraically, but include a graph of \(f^{\prime}\) to support your conclusion. b. Reasoning graphically, for what intervals of \(x\) -values is \(f^{\prime \prime}(x)>0\) ? What does this tell you about the behavior of the original function \(f ?\) Explain. c. Assuming that \(f(2)=-3\), estimate the value of \(f(1.88)\) by finding and using the tangent line approximation to \(f\) at \(x=2\). Is your estimate larger or smaller than the true value of \(f(1.88) ?\) Justify your answer.

Step by step solution

01

Find where f'(x) = 0

To find where the derivative is zero, solve the equation \( (x-1)e^{-x^2}=0 \). The factor \( e^{-x^2} \) is never zero for real values of \( x \) since the exponential function never equals zero. Thus, we only need to solve \( x-1=0 \), giving us \( x=1 \).

02

Confirm solution graphically

Graph the function \( f'(x)=(x-1)e^{-x^2} \). The graph will show that it intersects the x-axis at \( x=1 \), confirming that \( f'(x)=0 \) at \( x=1 \).

03

Find intervals where f''(x) > 0

To determine where \( f''(x) > 0 \), first find \( f''(x) \) by differentiating \( f'(x) \) again. Using the product rule, \[ f''(x) = e^{-x^2} + (x-1)(-2xe^{-x^2}) \] Simplify to get \[ f''(x) = e^{-x^2} (1 - 2x(x-1)) = e^{-x^2} (1-2x^2+2x) \] To determine the sign of \( f''(x) \), recognize that \( e^{-x^2} \) is always positive. Thus, solve \( 1-2x^2+2x > 0 \). From a sign analysis or graph, confirm \( f''(x) > 0 \) in the intervals.

04

Discuss behavior of f(x)

Where \( f''(x) > 0 \) implies the function \( f(x) \) is concave up. Thus, on these intervals, \( f(x) \) is increasing at an increasing rate.

05

Use tangent line approximation

To approximate \( f(1.88) \) using the tangent line at \( x=2 \), first find the equation of the tangent line. The slope is \( f'(2) = (2-1) e^{-4} = e^{-4} \), and the point is \( (2, -3) \). Thus, the tangent line is \( y + 3 = e^{-4} (x-2) \), or \( y = e^{-4} (x-2) - 3 \). To estimate \( f(1.88) \), substitute \( x=1.88 \) into the tangent line equation and solve.

06

Determine estimate accuracy

Since \( f''(2) > 0 \), the actual function is concave up. This means the tangent line approximation underestimates the actual value of \( f(1.88) \).

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Key Concepts

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

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate of change of a function with respect to a variable. Imagine you are driving a car and want to know how fast you are going at any given moment. The speedometer gives you the derivative of your position with respect to time. Mathematically, if you have a function \(y = f(x)\), its derivative, written as \(f'(x)\), tells you how \(y\) changes as \(x\) changes. In the given problem, the derivative \(f'(x) = (x-1)e^{-x^2}\) provides a lot of information about the function \(f\), even though we don't have the exact formula for \(f(x)\).

Concavity

Concavity refers to whether a curve is opening upwards or downwards. To determine this, we use the second derivative, \(f''(x)\). If \(f''(x) > 0\), the function is concave up, like a smile. If \(f''(x) < 0\), it's concave down, like a frown. This tells us how the rate of change itself is changing. For the given function, finding \(f''(x) = e^{-x^2} (1 - 2x(x-1))\) reveals intervals where the function \(f\) is concave up. These intervals give us insights into the behavior of \(f(x)\): it will be increasing at an increasing rate.

Tangent Line Approximation

Tangent line approximation is a handy tool for estimating the value of a function near a point, especially when the function itself is complex or unknown. The tangent line at \(x = a\) is the line that just 'touches' the curve at that point, with the same slope as the curve. It can be expressed as \(y = f'(a)(x - a) + f(a)\). For example, in our exercise, to estimate \(f(1.88)\) using the tangent line at \(x = 2\), we first find the tangent line equation: \(y = e^{-4}(x-2) - 3\). By plugging \(x = 1.88\) into this line, we get an estimate for \(f(1.88)\). Since the function is concave up at that point, our tangent line approximation will be an underestimate.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are essential for understanding the behavior of the function \(f(x)\). They could indicate local maxima, minima, or points of inflection. In our problem, solving \(f'(x) = (x-1)e^{-x^2} = 0\) gives us a critical point at \(x = 1\). To confirm if this is a maximum, minimum, or inflection point, we inspect the second derivative \(f''(x)\). Since the original exercise asks us to find intervals where \(f''(x) > 0\), we conclude that around the critical point, the function's behavior changes.