If
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
or
-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
Answer is a = 4, b = 2.
First derivative of an equation:
The derivative formula is one of the basic concepts used in calculus and the process of finding a derivative is known as differentiation. The derivative formula is defined for a variable 'x' having an exponent 'n'. The exponent 'n' can be an integer or a rational fraction. Hence, the formula to calculate the derivative is:
[tex]\frac{d}{dx} x^{n} = n x^{n-1}[/tex]
(Given)
[tex]f(x) = ax^{3} + bx^{2} - 5x + 9[/tex]
And f'(x) we get , [tex]f'(x) = 3ax^{2} + 2bx-5[/tex]
Now, also given f(-1) = 12
=> [tex]f(-1) =a (-1)^{3} + b(-1)^{2} - 5 (-1) + 9[/tex]
=> 12 = -a + b +5 + 9
=> a = b + 2 ------( 1 )
Also given, f'(-1) = 3.
i.e., [tex]3 = 3a(-1)^{2} + 2b(-1) - 5[/tex]
3 = 3a -2b -5
From eq - (1) we can replace a.
3 = 3(b + 2) - 2b - 5
3 = 3b + 6 - 2b - 5
b = 2.
a = 4.
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