calc multiple choice quiz answers + explanations

The graph of y=3x⁴-16x³+24x²+48 is concave down for...?

E → 2/3<x<2 - find f"(x) = 36x²-96x+48 = 12(3x-2)(x-2) - find that x=2/3 and x=2 - do the sign test - find that it's only negative between 2/3 and 2, so the answer is 2/3<x<2

At what values of x does f(x)=3x⁵-5x³+15 have a relative maximum?

A → -1 only - find f'(x) - f'(x) = 15x⁴-15x² = 15x²(x²-1) = 15x²(x-1)(x+1) - set f'(x) and get x=0, x=-1, and x=1 - do sign test - the only max is -1, so that must be the answer

The absolute maximum value of f(x)=x³-3x²+12 on the closed interval [-2,4] occurs at x=?

A → 4 - find f'(x) - f'(x) = 3x²-6x = 3x(x-2) - do sign test that includes the interval values - the only maximums are 0 and 4, but 0 is only a relative maximum in that interval, so the answer is 4

(calculator) The graphs of f', g', and h' are shown above. Which of the functions f, g, or h have a relative maximum on the open interval a<x<b?

A → f only - a change of the first derivative from pos to neg indicates a relative maximum - look at which graphs for which function changes from pos to neg at some point - f' is the only one that changes from pos to neg at some point

(calculator) Let f be the function given by f(x)=cos(2x)+ln(3x). What is the least value of x at which the graph of f changes concavity?

B → 0.93 - find f"(x) for the given function and then graph f"(x) on yor calculator - f changes concavity wherever f"(x) changes from pos to neg or from neg to pos - for this situation, that should occur at 0.93

The graph of y=f(x) is shown in the figure above. On which of the following intervals are f'(x)>0 and f"(x)<0?

B → II only (b<x<c) - you can tell that f'(x)>0 on the interval a<x<c because that's where f(x) is increasing - within the interval a<x<c, it is only concave down on the interval b<x<c, which indicates that f"(x)<0 - so the only interval where f'(x)>0 and f"(x)<0 is b<x<c

(calculator) f' is given by f'(x) = (cos²x)/(x) - 1/5. How many critical values does f have on the open interval (0,10)?

B → Three - find f'(x)=0 or does not exist - you get x=0, x=2.3, and x=3.7 - that's 3 critical points

If f"(x)=x(x+1)(x-2)², then the graph of f has inflection points when x=?

C → -1 and 0 only - find that x=0, x=-1, and x=2 - do the sign test with 0, -1, and 2 - inflection points occur when f"(x) change from pos to neg or from neg to pos - changes from pos to neg over -1, neg to pos over 0, and pos to pos over 2 - so the answer is -1 and 0 only

(calculator) If f' is given by f'(x)=eˣ-3x², at which of the following values of x does f have a relative maximum value?

C → 0.91 - graph the function f'(x) on your calculator - find the point where f'(x) changes from negative to positive - that should occur at 0.91

(calculator) (a) - For what values of x, -3<x<3, does f have a relative maximum, A relative minimum? Justify you answer. (b) - For what values of x is the graph of f concave up? Justify your answer. (c) Use the information found in parts (a) and (b) and the fact that f(-3)=0 to sketch a possible graph of f on the axes provided below.

(a) - there is a relative maximum at -2 because f'(x) crosses the x axis from pos to neg at that point - there is a relative minimum at 0 because f'(x) crosses the x axis from pos to neg at that point (b) - I did part c first because it was easier to tell which parts were concave up by looking at it on the graph of f(x) (c) - the points on f'(x) that are above the x axis are where f(x) is increasing - the points on f'(x) that are below the x axis are where f(x) is decreasing - by using the critical points that are given, you can draw the graph of f(x)

If c is the number that satisfies the conclusion of the Mean Value Theorem for f(x)=x³-2x² on the interval 0≤x≤2, then c=?

D → 4/3 - find f'(x) - you get f'(x) = 3x²-4x = 3x(x-4/3) - set f'(x)=0, and you get x=0 and x=4/3 - if c was 0, some answers would get left out, so it must be c=4/3

The graph of the derivative of f is shown in the figure above. Which of the following could be the graph of f?

B - the points on f'(x) that are above the x axis are where f(x) is increasing - the points on f'(x) that are below the x axis are where f(x) is decreasing - by using the critical points -1 and 1 that are given, you can draw the graph of f, and it should match the graph given by option B

The function f given by f(x)=3x⁵-4x³-3x has a relative maximum at x=?

A → -1 - find f'(x) = 15x⁴-12x²-3 = 3(5x⁴-4x²-1) - set f'(x)=0 and find that x=1 and x=-1 - do the sign test - find that between -1 and 1 is neg, and on either side is pos - with that info, you can see that -1 is the max

The graph of f', the derivative of f, is shown in the figure above. Which of the following describes all relative extreme of f on the open interval (a,b)?

A → One relative maximum and two relative minimums - draw the graph of f using the graph of f' - you will see that there is one relative maximum and two relative minimums

What is the x-coordinate of the point of inflection on the graph of y=1/3x³+5x²+24?

D → -5 - find f'(x) = x²+10x = x(x+10) - set f'(x)=0 and find that x=0 and x=-10 - do the sign test - it goes pos to neg over -10 and neg to pos over 0, so the inflection point is right in the middle, which would be -5

The graph of f is shown in the figure above. Which of the following could be the graph of f'?

A - the points on f(x) that are increasing are where f'(x) is above the x axis - the points on f(x) that are decreasing are where f'(x) is below the x axis - by using the critical points a and b that are given, you can draw the graph of f', and it should match the graph given by option A

A polynomial p(x) has a relative maximum at (-2,4), a relative minimum at (1,1), a relative maximum at (5,7), and no other critical points. How many real zeros does p(x) have?

B → Two - draw graph, and extend both ends down to cross the x-axis - real zeros occur where the graph crosses the x-axis, and in this graph, it crosses the x-axis two times, so the answer is two

An equation of the line tangent to y=x³+3x²+2 at its point of inflection is...?

B → y=-3x+1 - find y" and set equal to 0 - you get x=-1 - plug -1 into original equation for x to get y value - inflection point is (-1,4) - plug -1 into y' to get slope - slope is -3 - solve for b with 4=-3(-1)+b - b=1 - equation is y=-3x+1

(calculator) The graph of the function y=x³+6x²+7x-2cosx changes concavity at x=?

D → -1.89 - graph the function y on your calculator - find the values of the relative maximum and the relative minimum, and you will be able to find the value of the point in the middle, which is the inflection point

The graph of a twice-differentiable function f is shown in the figure above. Which of the following is true?

D → f"(1) < f(1) < f'(x) - you can clearly see that f(1)=0 - the tangent line of f(1) is positive so f'(x)>0 - f"(1)<0 because the function is concave down - so the answer is f"(1) < f(1) < f'(1)

If the graph of y=x³+ax²+bx-4 has a point of inflection at (1,-6), what is the value of b?

B → 0 - find f" - set f"(1)=0 - solve for a - you get a=-3 - plug -3 for a, 1 for x, and -6 for Yyinto original equation to find b - you get b=0

If g is a differentiable function such that g(x)<0 for all real numbers x and if f'(x)=(x²-4)g(x), which of the following is true?

B → f has a relative minimum at x=-2 and a relative maximum at x=2 - simplify f'(x) to f'(x)=(x-2)(x+2)g(x) - do the sign test; g(x) will always be negative - you should get neg to pos over -2, and pos to neg over 2 - so x=-2 is a relative minimum and x=2 is a relative maximum

If f is the function defined by f(x)=3x⁵-5x⁴, what are all the x-coordinates of points of inflection for the graph of f?

C →1 - find f"(x) - f"(x) = 60x³-60x² = 60x²(x-1) - find where f"(x) changes from pos to neg or from neg to pos by using the sign test - goes from neg to neg for 0, and neg to pos for 1, so the answer is just 1

What are all values of x for which the function f defined by f(x)=x³+3x²-9x+7 is increasing?

C →x<-3 or x>1 - find f'(x) = 3x²+6x-9 = 3(x-1)(x+3) - set f'(x)=0 and find that x=1 and x=-3 - do the sign test - you will see that it's increasing to the left of -3 and to the right of 1

Which of the following pairs of graphs could represent the graph of a function and the graph of its derivative?

D → I and III - just draw the graphs of f' based on the graphs of f given - you should get that I and III are the same as the answers they give

The graph of f' is shown in the figure above. Which of the following could be the graph of f?

E - the points on f'(x) that are above the x axis are where f(x) is increasing - the points on f'(x) that are below the x axis are where f(x) is decreasing - by using the critical points -2 and 2 that are given, you can draw the graph of f, and it should match the graph given by option E

The graph of y=h(x) is shown above. Which of the following could be the graph of y=h'(x)?

E - the points on h(x) that are increasing are where h'(x) is above the x axis - the points on h(x) that are decreasing are where h'(x) is below the x axis - by using the critical points that you can draw where h(x) changes from pos to neg or from neg to pos, you can draw the graph of h'(x) - it should match the graph given by option E