   Chapter 3.5, Problem 42E

Chapter
Section
Textbook Problem

# In the theory of relativity, the energy of a particle is E = m ο 2 c 2 + h 2 c 2 / λ 2 where m ο is the rest mass of the particle, λ is its wave length, and h is Planck’s constant. Sketch the graph of E as a function of λ . What does the graph say about the energy?

To determine

To sketch:

The curve of y

Explanation

1) Concept:

i) The domain is the set of x values for which the function is defined.

ii) To find x-intercept, put y=0, and to find y-intercept, put x=0 in the given function.

iii) Symmetry: To find symmetry, replace x by –x and check the behaviour of function. Thus, if f-x=fx, then it is an even function, so its graph is symmetric about the y-axis. If f-x=-fx, then it is an odd function, so its graph is symmetric about the origin. And if f-x-fxfx, then it has no symmetry.

iv) An asymptote is a tangent at infinity. To find horizontal, vertical, and slant asymptote, follow the rules.

v) A function is increasing if f'x>0  and decreasing if f'x<0 in that particular interval.

vi) The number f(c) is a local maximum value of f  if fcf(x) when x is near c and is a local minimum value of f if fc f(x) when x is near c.

vii) If f''x>0, the function is concave up and if f''x<0, the function is concave down in that particular interval. And f''x=0, gives the values of inflection points

2) Given:

E=f(λ)=m02c4+h2c2λ2

3) Calculations:

Here, first find the domain of the given function and the x & y intercepts. Next, check the symmetry, asymptotes, intervals of increase and decrease, local maximum and minimum values, concavity, and points of inflection. Using these, sketch the curve

A. Domain:

So, the domain is (0, ), since λ can only be positive

B. Intercepts:

There are no x intercepts and y intercepts

C

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