Calculus I for Programmers

Difference quotients, by hand

Calculus I for Programmers

Difference quotients, by hand

Medium0 of 6 parts

Each of these is a difference quotient. Simplify the top until every term carries an hh, cancel the hh in the denominator, and then answer the question asked. No calculator will help with any of them, and no exam will let you use one.

f(a+h)−f(a)h\frac{f(a+h) - f(a)}{h}
The quantity being simplified in every part below.

Where the answer is in terms of aa, write it using the letter aa. Where it is in terms of both aa and hh, use both.

1

Simplify (a+h)2−a2h\dfrac{(a+h)^2 - a^2}{h} for h≠0h \neq 0. Give the answer in terms of aa and hh.

in terms of a, h

2

Now take it to the limit: lim⁡h→0(a+h)2−a2h\lim_{h \to 0} \dfrac{(a+h)^2 - a^2}{h}, in terms of aa.

in terms of a

3

f(x)=3x2−5xf(x) = 3x^2 - 5x. Simplify f(a+h)−f(a)h\dfrac{f(a+h) - f(a)}{h}, in terms of aa and hh.

in terms of a, h

4

f(x)=1xf(x) = \dfrac{1}{x}. Find lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}, in terms of aa.

in terms of a

5

f(x)=xf(x) = \sqrt{x}. Find lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}, for a>0a > 0, in terms of aa.

in terms of a

6

f(x)=7f(x) = 7 for every xx. Find lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}.