AP Calculus 1992 BC4 problem!

Test Preparation
1

<p>Could someone pleease be as kind as to solve this problem? It's giving me quite a bit of trouble. Also, if you have the answers to the 1993 AB1 question for Calc, could you please send it to me? I've solved it but I am not sure if my answer is correct. Help EXTREEEMELY appreciated. </p>

<p>Let f be a function by the piece-wise function f(x) = 2x-x^2, for x is less than or equal to 1
x^2 + kx+ p, for x > 1</p>

<p>a. For what values of k and p will f be continuous and differentiable at x = 1</p>

<p>b. For what values of k and p found in part a, on what interval or intervals is f increasing?</p>

<p>c. Using the values of k and p found in part a, find all points of inflection of the graph of f. Support your conclusion.</p>

2

<p>starryq2261 - there is a better place for the non-SAT/ACT questions:
<a href="http://physicsforums.com/%5B/url%5D"&gt;http://physicsforums.com/&lt;/a>
Caveat: you'll have to show your attempts at solving. :(</p>

<p>I. Without calculus.
a.
g(x) = 2x-x^2 = -(x-1)^2 + 1
Zeroes x1=0 and x2=2, vertex (1, 1)</p>

<p>g(x) and h(x) = x^2 + kx+ p are both parabolas, so
for h(x) = x^2 + kx+ p to "smoothly" continue g(x) for x>1 h(x) should have the same vertex as g(x) and be the same graph opening up.
h(x) = (x-1)^2 + 1 = x^2 - 2x +2 = x^2 + kx+ p
k = -2, p = 2</p>

<p>f(x) = g(x) for x <= 1
f(x) = h(x) for x > 1.
Straight lines tangent to g(x) at x=a, a<=1, "smoothly" become tangent to h(x) for a going past x = 1 , so f(x) is differential at x = 1.</p>

<p>b.
Parabola g(x) opens down, left of vertex at x<=1 it's increasing,
parabola h(x) opens down, right of vertex at x<=1 it's increasing as well;
f(x) is increasing on R ]-∞, +∞[.</p>

<p>c.
Both graphs g(x) and h(x) are tangent to y=1, but are on opposite sides of it:</p>

<h1>(1, 1) is a point of inflection.</h1>

<p>II. Calculus.
a.
g(x) = 2x-x^2
h(x) = x^2 + kx+ p</p>

<p>g'(x) = 2 - 2x
h'(x) = 2x + k</p>

<p>f(x) is differential at x = 1,
g'(1) = h'(1)
0 = 2 + k,
k = -2</p>

<p>f(x) is continuous at x = 1,
g(1) = h(1)
2*1 - 1^2 = 1^2 + kx + p
1 = 1 - 2 + p
p = 2
h(x) = x^2 - 2x + 2</p>

<p>f(x) = g(x) for x <= 1
f(x) = h(x) for x > 1.</p>

<p>b.
f'(x) = 2 - 2x > 0 for x < 1
f'(x) = 2 - 2x = 0 for x = 1
f'(x) = 2x - 2 > 0 for x > 1
f(x) is increasing on R ]-∞, +∞[.</p>

<p>c.
f"(x) = -2 for x <= 1.
f"(x) = 2 for x > 1
so f"(x) changes sign at x = 1 only and
(1, 1) is the point of inflection.
It's the only inflection point because "normal" parabolas don't have them.
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