We're looking for a polynomial
p(x) with these properties:
It has a local max at
x=1 and a local
min at
x=3.
p(1)=6‌ and ‌p(3)=2We want to find the value of its derivative at
x=0, which is
p′(0).
Solution
Since
p(x) has a local maximum and a local minimum, it must have at least two turning points. This means the derivative
p′(x) must have at least two roots. Therefore, the polynomial
p(x) must be at least a cubic (degree 3) polynomial.
Formulate the Derivative:
Since
p(x) has a local max at
x=1 and a local min at
x=3, we know that
p′(1)=0 and
p′(3)=0. Therefore, we can express
p′(x) in the form:
p′(x)=k(x−1)(x−3)where
k is some constant.
Find the Polynomial:
Integrate
p′(x) to find
p(x) :
‌p(x)=∫p′(x)‌dx=∫k(x−1)(x−3)‌dx=∫k(x2−4x+3)‌dx
‌p(x)=k(‌−2x2+3x)+Cwhere
C is the constant of integration.
Use the Given Conditions:
We know that
p(1)=6 and
p(3)=2. Plug these values into the equation for
p(x) :
Now we know
C=2. Substitute this back into the first equation:
‌k(‌)+2=6‌k(‌)=4‌k=3 Complete the Polynomial:
Substitute the values of
k and
C back into the equation for
p(x) :
Find the Derivative:
Differentiate
p(x) to find
p′(x) :
p′(x)=3x2−12x+9 Calculate
p′(0) :
Substitute
x=0 into
p′(x) :
p′(0)=3(0)2−12(0)+9=9Answer:
Therefore,
p′(0)=9. So, the correct option is B .