It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
Apply the reciprocal rule: If f(x) = 1/h(x) then f'(x) = -h'(x)/(h(x))^2
24
The derivative of f(x) is lim h-->0 [f(x+h)-f(x)]/h. So let f(x) = -5x. The derivative is lim h-->0 [-5(x+h)- -5(x)]/h = lim h-->0 [-5x - 5h + 5x]/h = lim h-->0 -5h/h Since the limit h-->0 of h/h is 1, the derivative is -5
h = -5 -7 * -5 = 35 A negative times a negative equals a positive.
f = 54
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
Not according to the usual definitions of "differentiable" and "continuous".Suppose that the function f is differentiable at the point x = a.Then f(a) is defined andlimit (h -> 0) [f(a+h) - f(a)]/h exists (has a finite value).If this limit exists, then it follows thatlimit (h -> 0) [f(a+h) - f(a)] exists and equals 0.Hence limit (h -> 0) f(a+h) exists and equals f(a).Therefore f is continuous at x = a.
Means that h equals 1660.
0.5
5 yards and a first down for Dante Hall ? Devin Hester?
24
Apply the reciprocal rule: If f(x) = 1/h(x) then f'(x) = -h'(x)/(h(x))^2
The derivative of f(x) is lim h-->0 [f(x+h)-f(x)]/h. So let f(x) = -5x. The derivative is lim h-->0 [-5(x+h)- -5(x)]/h = lim h-->0 [-5x - 5h + 5x]/h = lim h-->0 -5h/h Since the limit h-->0 of h/h is 1, the derivative is -5
If f(x)=x2+10, then f(x+h)=?f(x+h)=(x+h)2+10 (since f(x)=x2+10, substitute the x in x2 to (x+h)2)=(x+h)(x+h)+10 (then multiply (x+h) by (x+h) by doing the FOIL method)=x2+xh+xh+h2+10 (First: x*x, Outside: x*h, Inside: h*x, Last: h*h)=x2+2xh+h2+10 (combine like terms (xh+xh=2xh))So if f(x)=x2+10, then f(x+h)=x2+2xh+h2+10
Ten is the number of fingers on two hands.
h = -5 -7 * -5 = 35 A negative times a negative equals a positive.