critical numbers and second derivatives?
For the function, find all critical numbers and then use the second-derivative test to determine whether the function has a relative maximum or minimum at each critical number. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
f(x) = x + 16/x
relative maxima x =
relative minima x =
- Ray SLv 73 days agoFavorite Answer
f(x) = x + 16/x
f(x) = x + 16x⁻¹
f '(x) = 1 - 16x⁻² ⇒ f '(x) = (1 + ⁴∕ₓ)(1 - ⁴∕ₓ) = 0 f or x = ±4 ← Critical Numbers
f ''(x) = 0 - (-32x⁻³)
f ''(x) = 32/x³ ANSWER
f ''(4) = 32/4³ > 0 ⇒ concave up ⇒ ⇒ relative minimum at x = 4
f ''(-4) = 32/(-4)³ < 0 ⇒ concave down ⇒ relative maximum at x = -4
- lenpol7Lv 72 days ago
Let f(x) = y
Hence y = x + 16/x = x + 16x^-1
Make the first differential and equate to zero. This will gave the local max/min.
dy/dx = 1 + (-1)16x^(-2)
1 - 16x^-2 = 0
( 1- 4/x)(1 + 4/x) = 0
1 - 4/x = 0
4x = 1
x = 1/4
1 + 4/x = 0
x = -1/4
We don't know if '1/4' or '-1/4'. So we differentiate again. If the result is positive/negative then the curve is at a local Min/Max.
d2y/dx2(1 - 16x^-2) = 0 - (-2)(16)x^-3 = 32/x^3
When x = 1/4 = the 32/( 1/4)^3 = (+)2048 positive result ,so cure is at a local minima.
When x = -1/4 - the 32/(-1/4)^3 = (-)2048 negative result, so curve is at a local max.
- ArgentLv 73 days ago
f(x) = x + 16/x = x + 16xˉ¹.
f '(x) = 1 + (-16)xˉ² = 1 - 16xˉ².
From the Wikipedia article on "critical point", we have: "a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero."
f '(x) = 0 → 1 - 16xˉ² = 0, or 1 = 16xˉ², or xˉ² = 1/16, or x² = 16, so x = ±4.
x=-4 and x=4 are two of the critical points. (The other critical point is at x=0, where the function has a discontinuity and is therefore not differentiable.)
To find the max and min values, take the second derivative:
f ''(x) = (-2)(-16xˉ³) = 32xˉ³.
At x=-4, f ''(x) = -2048, which is < 0, so the point is a relative maximum.
At x=4, f ''(x) = 2048, which is > 0, so the point is a relative minimum.
At x=0, there is no first derivative, so there is no second derivative either. (The function has no (finite) value at all, there, so there is no point to evaluate for max or min.)
- alexLv 73 days ago
Solve f '(x) = 0 to find the x-coordinate of the critical number(s)