Asked by Ashton
(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.)
Answers
Answered by
Ashton
(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).
(enter increasing, decreasing, or neither).
Is it concave up or concave down?
(enter up, down, or neither).
Answered by
Steve
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
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