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mash [69]
4 years ago
12

Find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 3 at x = −2 and a local minimum value of 0 at

x = 1. f(x) =
Mathematics
2 answers:
zheka24 [161]4 years ago
8 0

Answer:

Step-by-step explanation:

f(-2)=3

f(1) =0

f'(-2)=0

f' (1) = 0 ..... continue

Helen [10]4 years ago
3 0

f(x)=ax^3+bx^2+cx+d

\implies f'(x)=3ax^2+2bx+c

Local extrema occur wherever f'(x)=0.

We want

  • a local max of 3 when x=-2, which means

f(-2)=-8a+4b-2c+d=3

f'(-2)=12a-4b=0\implies3a=b

and

  • a local min of 0 when x=1, which means

f(1)=a+b+c+d=0

f'(1)=3a+2b+c=0

Solve the system for a,b,c,d; you should get

a=\dfrac19,b=\dfrac13,c=-1,d=\dfrac59

so that the cubic we're looking for is

\boxed{f(x)=\dfrac{x^3}9+\dfrac{x^2}3-x+\dfrac59}

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