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show that there exists some value ‘c’ on [a, b] such that f(c) = 2
f is continuous because… (type of function? differentiable? given?)
h(a) < 2 and h(b) >2
thus, according to the ivt, there exists some value ‘c’ on [a, b] so that f(c) = 2
show that f(x) meets conditions for the mvt then find some value ‘c’ guaranteed by the theorem on [a, b]
f is differentiable because…
[f(b) - f(a)]/[b - a] = f’(c)
find intervals of increase/decrease and give values for any extreme for f(x)
f is increasing on [closed interval] as f’(x) > 0
f is decreasing on [closed interval] as f’(x) < 0
find an absolute minimum and maximum for f(x)
f has an absolute maximum of [y-value] when [x-value]
f has an absolute minimum of [y-value] when [x-value]
find a local minimum and maximum for f(x)
f has a local minimum when [x-value] as f’(x) changes from negative to positive about [x-value]
f has a local maximum when [x-value] as f’(x) changes from positive to negative about [x-value]
find when f(x) is concave up and concave down
f is concave up from (open interval) as f’’(x) > 0
f is concave down from (open interval) as f’’(x) < 0
f has an inflection point when [x-value] as f’’(x) changes signs about [x-value]