(1 point) Is the graph of y=sin(x4) increasing or decreasing when x=14?
(enter increasing, decreasing, or neither).
Is it concave up or concave down?
(enter up, down, or neither).
(1 point) Find the inflection points of f(x)=4x4+22x3−18x2+10. (Give your answers as a comma separated list, e.g., 3,-2.)
2 answers
ok, I'll help on this one too. But I'm sure your text explains this all in detail.
an inflection point occurs (usually) when f"(x) = 0 and f'(x) ≠ 0
the graph is concave up if f"(x) > 0
concave down if f"(x) < 0
for y = 4x^4+22x^3−18x^2+10
y' = 16x^3+66x^2-36x = 2(8x^3+33x^2-18x)
y" = 2(24x^2+66x-18) = 12(4x^2+11x-3) = 12(x+3)(4x-1)
so, there are inflection points at -3, 1/4 since f' ≠ 0 there
for y = sin(x^4)
y' = 4x^3 cos(x^4)
y" = 4x^2(3cos(x^4) - 4x^4 sin(x^4))
y'(14) = 4*14^3 cos(14^4) > 0 so f is increasing
y"(14) = 4*14^2(3cos(14^4) - 4*14^4 sin(14^4)) < 0 so the graph is concave down
an inflection point occurs (usually) when f"(x) = 0 and f'(x) ≠ 0
the graph is concave up if f"(x) > 0
concave down if f"(x) < 0
for y = 4x^4+22x^3−18x^2+10
y' = 16x^3+66x^2-36x = 2(8x^3+33x^2-18x)
y" = 2(24x^2+66x-18) = 12(4x^2+11x-3) = 12(x+3)(4x-1)
so, there are inflection points at -3, 1/4 since f' ≠ 0 there
for y = sin(x^4)
y' = 4x^3 cos(x^4)
y" = 4x^2(3cos(x^4) - 4x^4 sin(x^4))
y'(14) = 4*14^3 cos(14^4) > 0 so f is increasing
y"(14) = 4*14^2(3cos(14^4) - 4*14^4 sin(14^4)) < 0 so the graph is concave down
1 answer