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Given the following information about p(x) find a and b. Then use to find slopes in interval [0,16]?
Let p(x) be continuous and differentiable.
p(x) = 16a/sqrt(x) when x > 4
p(x) = bx^2 + 2x + 56 when x </= 4
Prove that a = and b = .
Then, using those values of a and b find the value(s) of x in the interval [0,16] at which the instantaneous slope equals the average slope.
1 Answer
- anonimousLv 68 years agoFavorite Answer
To be continuous
When x=4
16*a/sqrt(x)=b*x^2 +2*x +56
16*a/2=b*16+8+56
8*a=b*16+64
a=2*b+8
for the differential to be continuous
When x=4
16*a*(-1/2)*x^(-3/2)=b*2*x+2
-8*a/8=8*b+2
a=-8*b-2
subtracting the simplified equation for differentiable from the simplified equation for continuous eliminates the variable a
a-a=2*b+8-(-8*b-2)
0=10*b +10
b=-1
a=2*b+8=-2+8=6
a=-8*b-2=8-2=6
The average slope is not a well defined expression. I will use as my definition the sum of the instantaneous slopes at the end points divided by two.
p'(0)=b*2*0+2=2
p'(16)=16*a*(-1/2)*16^(-3/2)=-8*6/64=-3/4
{p'(0)+p'(16)}/2=(2-3/4)/2=5/8
-2*x+2=5/8
2*x=2-5/8
x=11/16 when the instantaneous slope is equal to the average slope over the range.
