using the derivative rules: y 8x4 9 4. Use differentials to approximate V9.3 usi
ID: 2864017 • Letter: U
Question
using the derivative rules: y 8x4 9 4. Use differentials to approximate V9.3 using the value at x of the function y Nx Round your answer to the nearest thousandth. 5. Find the equation of the tangent line to x2-2 xy- y 17 at the point (-3,2) 6. For the function y -4x +8x2, use the first and second derivative tests to (a) determine the intervals of increase and decrease. (b) determine the relative maxima and minima. (c) determine the intervals of concavity. (d) determine the points of inflection. (e) sketch the graph with the above information indicated on the graph 7. A box with a square base is open at the top. If 108 square feet of material are maximum volume possible for the box? 8. The radius r of a sphere is increasing at a rate of 4 inches per minute. Find th l ulu diameter is 12 inches. Hint: v Tr3Explanation / Answer
Solution:
To find the critical points, set the first derivative = 0 and solve for x.
f(x) = 4x^3 + 8x^2
f'(x) = 12x^2 + 16x = 0
4x(3x + 4) = 0
x = 0, -4/3 (these are the critical points or values)
to find max/min, use the second derivative test.
f''(x) = 24x + 16
now plug in each critical point into the second derivative.
if you get a positive answer, the point is a relative minimum.
if you get a negative answer, the point is a relative maximum.
f''(0) > 0, therefore, x = 0 is a relative minimum
f''(-4/3) < 0, therfore x = -4/3 is a relative maximum
To find where the function is increasing and decreasing, use numberline analysis. draw a numberline and plot the 2 critical points we have. then plug in a number from interval into the derivative to see where the derivative is positive or negative. x<-4/3 is negative so the function is decreasing on that interval. -4/3<x<0 is positive so the function is increasing on that interval.
To find points of inflection and where the graph is concave up and down, you need to find possible inflection points by setting the 2nd derivative = 0.
f''(x) = 24x + 16 = 0
x = -16/24 = -2/3 <=== inflection point (concavity changes)
f'' (0) = 16 <=== positive value x = 0 is a minimum (concave up)
f'' (-4/3) = -16 <=== Negative value x = -4/3 is a maximum (concave down)
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