f'(x)={f(x)-f(y)}/{x-y}
for the second one:
put (2,1) into f(x).
this is your first equation with a and b.
now,
find f'(x).
f'(x)=0
put x=1.
And there you have second equation.
solve system of equations finding a and b.
Hello!
I need some help please with two revision questions.
I read my Calculus book and I did not understand it!
Q1.
Suppose that 3<= f ' (x) <= 5 for all values of x. Determine the highest and lowest possible values for the expression A = f (8) - f (2).
Am I meant to use "MVT" ??
Q2.
For what values of the numbers a and b does the function f(x) = axe^bx²
have the maximum value f(2) = 1? Give reasons.
Sorry this I do not know.
That seems most reasonable to me (f(8)- f(2))/(8- 2)= (f(8)- f(2))/6 is equal to the derivative of f at some point- but you know that can never be larger than 5 or less than 3.
At a max or min, the derivative will be 0. . Where is that equal to 0? What is the value of f there? What must a and b equal in order that that result be 1?Q2.
For what values of the numbers a and b does the function f(x) = axe^bx²
have the maximum value f(2) = 1? Give reasons.
Sorry this I do not know.