Pre-Calc Questions

Question

Give a practical interpretation in words of the function
1) k(g(t)), where L=k(H) is the length of a steel bar at temperature H and H=g(t) is temperature at time t
2) t(f(H)), where t(v) is the time of a trip at velocity v, and v=f(H) is velocity at temperature H
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Find a simplified formula for the difference quotient --- (f(x+h)-f(x))/h
3) f(x)=x^2 +x
4) f(x)=sqrtx
5) f(x)= 1/x

Answer 1

Give a practical interpretation in words of the function

1) k(g(t)), where L=k(H) is the length of a steel bar at temperature H and H=g(t) is temperature at time t

The length of the steel bar is a function of its temperature.

2) t(f(H)), where t(v) is the time of a trip at velocity v, and v=f(H) is velocity at temperature H

The time of a trip is a function of the temperature.

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Find a simplified formula for the difference quotient --- (f(x+h)-f(x))/h

3) f(x) = x^2 + x

f(x+h) = (x+h)^2 + (x+h)

f(x+h) = x^2+2hx+h^2 + x+h
–f(x) = –x^2 – x
––––––––––––––––––––––––––
f(x+h)–f(x) = 2hx+h^2 +h

(f(x+h)–f(x))/h = 2x + h + 1

4) f(x) = √(x)

f(x+h) = √(x+h)
–f(x) = –√(x)
––––––––––––––––––
(f(x+h)–f(x))/h = (√(x+h)–√(x))/h = d

d = (√(x+h)–√(x))(√(x+h)+√(x)) / (h(√(x+h)+√(x)))

d = ((x+h)–x)/(h(√(x+h)+√(x)))

d = h/(h(√(x+h)+√(x)))

d = 1/(√(x+h)+√(x))

This is “simplified” because we can now let h get very small and find d:

h → 0 ==> d → 1/(√(x+0)+√(x)) = 1/(2 √(x))

In calculus this process is called "finding the derivative".

5) f(x) = 1/x

f(x+h) = 1/(x+h)
–f(x) = –1/x
––––––––––––––––
f(x+h)–f(x) = 1/(x+h) – 1/x = (x – (x+h))/(x(x+h))

(f(x+h)–f(x))/h = (–h/(x(x+h)))/h = –1/(x(x+h) = d

d = –1/(x(x+h)) is the answer because, in calculus:

h → 0 ==> d → –1/(x(x+0)) = –1/x^2