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3 years ago

For the function f(x)=ax^3+bx^2-5x+9, determine the values of a and b so that f(-1)=12 and f'(-1)=3

Mathematics

1 answer:

Alecsey [184]3 years ago

4 0

f(x) = ax ³ + bx ² - 5x + 9

then

f '(x) = 3ax ² + 2bx - 5

Given that f (-1) = 12 and f '(-1) = 3, we get the system of equations

-a + b + 5 + 9 = 12

3a - 2b - 5 = 3

-a + b = -2

3a - 2b = 8

Multiply through the first equation by 2 and add it to the second one to eliminate b and solve for a :

2(-a + b) + (3a - 2b) = 2(-2) + 8

-2a + 2b + 3a - 2b = -4 + 8

a = 4

Substitute this into the first equation above to solve for b :

-4 + b = -2

b = 2

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$\bf (x+1)(x-3)(x-2-4i)(x-2+4i)
\\\\
\textit{now, let's take a peek at }(x-2-4i)(x-2+4i)\\
\textit{and recall our }\textit{difference of squares}
\\ \quad \\
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-----------------------------\\\\
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\\\\\
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-----------------------------$

$\bf thus
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\begin{array}{llll}
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\end{array}\implies (x+1)(x-3)(x^2-2x+20)
\\\\\\
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\begin{array}{llll}
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\end{array}
\\\\\\
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