Asked by Jay
For each function, the point given is the maximum or minimum. Use the difference quotient to verify that the slope of the tangent at this point is zero.
a) f(x) = 0.5x^2 + 6x + 7.5; (-6, -10.5)
Difference quotient is
f(a + h) - f(a)/h
m= f(a+h) - f(a)/h
= f(-6+h) - f(-6)/h
= 0.5(-6+h)^2 + 6(-6+h)+ 7 - (-10.5)/h
What do I do next?
a) f(x) = 0.5x^2 + 6x + 7.5; (-6, -10.5)
Difference quotient is
f(a + h) - f(a)/h
m= f(a+h) - f(a)/h
= f(-6+h) - f(-6)/h
= 0.5(-6+h)^2 + 6(-6+h)+ 7 - (-10.5)/h
What do I do next?
Answers
Answered by
Jay
= 0.5(-6+h)^2 + 6(-6+h)+ 7.5 - (-10.5)/h *
Answered by
Reiny
just go ahead and work it out ...
m = [ .5(36 - 12h + h^2) - 36 + 6h + 7.5 + 10.5]/h
= [ 18 - 6h + h^2/2 - 36 + 6h + 7.5 + 10.5]/h
= (h^2/2)/ h
= h/2
now as h ---> 0 , m = 0
m = [ .5(36 - 12h + h^2) - 36 + 6h + 7.5 + 10.5]/h
= [ 18 - 6h + h^2/2 - 36 + 6h + 7.5 + 10.5]/h
= (h^2/2)/ h
= h/2
now as h ---> 0 , m = 0
Answered by
Jay
how did you get h^2/2?
Answered by
Jay
somebody? :|
Answered by
Reiny
.5 is the same as 1/2, so
.5(h^2) = (1/2)h^2 = h^2/2
.5(h^2) = (1/2)h^2 = h^2/2
Answered by
Jay
Oh wow thanks~