Calc AB Questions

Question

1. If h(x) is a twice differentiable function such that h(x) < 0 for all values of x, then at what value(s) does the graph of g(x) have a relative maximum if g'(x) = h(x) × (9-x²)?

A. x = 3 and x = -3 B. x = -3 only C. x = 9 only D. x = 3 only E. g(x) = has no relative max

1. An equation of the line tangent to the curve y = x³-6x² at its point of inflection is

A. y = -12x+12 B. y = -12x+40 C. y = 12x-40 D. y = -12x+8 E. y = 12x-8

1. If f(x) = cosx, then the Mean Value Theorem guarantees that somewhere between o and π/3, f(x) =

A. π/6 B. -π/6 C. -3/2π D. 3/2π E. 1/2

Answer 1

Problem 1

First of all, g(x) is defined everywhere because it is the integral of

h(x)×(9 - x²).

Therefore relative maximum occurs at any point where

g'(x) = 0 and g"(x) < 0.

First find all x satisfying

g'(x) = 0

h(x)×(9 - x²) = 0

A product is 0 iff either operand is 0. Since never h(x) = 0, we are left with

9 - x² = 0

x = ±3

Now for the condition

g"(x) < 0

h'(x)×(9 - x²) + h(x)(-2x) < 0

Substituting the two critical points ±3 we see that only x = -3

satisfies the condition, as h(x) < 0.

Problem 2

Inflection point occurs where the second derivative is 0

y' = 3x² - 12x

y" = 6x - 12

Therefore

y" = 0 at x = 2

At x = 2, y = -16, y' = -12.

Therefore the tangent is of the form y = -12x + b for some b.

The value b is determined from the condition that it passes through the point

(2, -16):

-16 = -12×2 + b

b = 8

So the tangent is y = -12x + 8

Problem 3.

f(0) = 1, f(π/3) = 1/2.

The slope of the line connecting the points (0, 1) and (π/3, 1/2) is

-3/2π.

The Mean Value Theorem guarantees there there is a point x where

f'(x) = -3/2π.

I think there is a typo in the question:

The Mean Value Theorem does not guarantee anything about the value of f itself.