Given: F(x)= 5x^2
a) find a simplified form of the difference quotient
b) find the average rate of change (difference quotient)x=4, h=2
Please help...
http://www.mathwords.com/d/difference_quotient.htm
Ok I went to this site and I saw the formula, how do I set up a problem like x^2-1 using the formula?
To find the difference quotient for the function F(x) = x^2 - 1 using the formula, you need to substitute the given function and simplify. The difference quotient is defined as:
[f(x + h) - f(x)] / h
a) Difference quotient for F(x) = 5x^2:
To find the difference quotient for F(x) = 5x^2, substitute the function into the formula:
[f(x + h) - f(x)] / h = [5(x + h)^2 - 5x^2] / h
Now, simplify the expression by expanding and simplifying:
[5(x^2 + 2hx + h^2) - 5x^2] / h = (5x^2 + 10hx + 5h^2 - 5x^2) / h
= (10hx + 5h^2) / h
= 10x + 5h
So, the simplified form of the difference quotient for F(x) = 5x^2 is 10x + 5h.
b) Average rate of change when x = 4 and h = 2:
To find the average rate of change, substitute the values of x = 4 and h = 2 into the simplified form of the difference quotient:
Average rate of change = 10x + 5h
= 10(4) + 5(2)
= 40 + 10
= 50