Question

without using graphs as an aid, use your knowledge of differentials to discuss the continuity of the function g(x)= 5x^3=10 at point x=4

Answer 1

http://www.youtube.com/watch?v=dQw4w9WgXcQ this will help

Answer 2

5x^3=10 at point x=4 ... this doesn't work, sorry?
5x^3=10
therefore x = a constant... = cube root of 2... so x never equals 4

Answer 3

Sorry I meant 5x^3-19 at point x=4

Answer 4

my first thing is always to "solve" for y:
g(x) = 5x^3-19
g(4) = 5(4)^3-19
g(4) = 320-19
g(4) = 301

so at x = 4, y = 301

now, slope of the line (first derivative):
g(x) = 5x^3-19
g'(x) = 15x^2

so at x = 4
g'(4) = 15(4)^2
g'(4) = 15 * 16
g'(4) = 240

therefore: slope is positive and crazy steep!

now: is slope increasing or decreasing (second derivative):
g(x) = 5x^3-19
g'(x) = 15x^2
g''(x) = 30x

so at x = 4
g''(x) = 30x
g''(4) = 30(4)
g''(x) = 120

slope of the line is increasing at this point, and sharply increasing that is...

Answer 5

ok, so my question then is for the actual answer, what is it? I have already figured out most of what you had written, I do not understand what to put down as an answer

Answer 6

the question just says discuss the continuity of this function around the point 4, so not really sure, but would go with something like:
function is positive in the cartesion plane at this point,
function's slope is steep and positive at this point
function continues to get steeper continuing past this point as x increases