polynomial with derivative (1 Viewer)

juantheron · Apr 8, 2012

Let $p(x)$ be a real polynomials of Least Degree Which has Local Maximum at $x=1$ and Local Minimum

at $x=3$ .If $p(1)=6$ and $p(3)=2$ . Then $p^{'}(0)=$

Carrotsticks · Apr 8, 2012

The polynomial of least degree having 2 distinct local max/min points is a cubic polynomial, which can be generalised to become:

$P(x) = ax^3 + bx^2 + cx + d \Rightarrow P'(x) = 3ax^2 + 2bx + c$

We will now use the details that they gave us to find the values of a, b, c and d:

Local maximum at x=1:

$P'(1) = 0 \Rightarrow 3a+2b+c=0 \qquad (1)$

Local minimum at x=3:

$P'(3) = 0 \Rightarrow 27a+6b+c=0 \qquad (2)$

(1) - (2) to eliminate 'c' gives us:

$24a + 4b = 0 \Rightarrow 6a + b = 0 \qquad (3)$

Now we use P(1) = 6:

$a + b + c + d = 6 \qquad (4)$

Similarly for P(3) = 2:

$27a + 9b + 3c + d = 2 \qquad (5)$

Solve simultaneously to find the value of 'c' because that is the value that P'(0) takes ie:

$P'(0) = c$

juantheron · Apr 8, 2012

Thanks Carrotsticks , after seeing Your Solution, I get This Method

$\hspace{-16}$Let $\bf{\frac{d}{dx}\{p(x)\}=a(x-1).(x-3).................(1)}$\\\\ So $\bf{\frac{d}{dx}\{p(x)\}=a(x^2-4x+3)=ax^2-4ax+3a}$\\\\ So $\bf{p(x)=\frac{ax^2}{3}-2ax^2+3ax+d}$\\\\ Put $\bf{x=1}$\\\\ $\bf{4a+3d=18}$\\\\ Again Put $\bf{x=3}$\\\\ $\bf{d=2}$\\\\ So We Get $\bf{a=3}$\\\\ Now $\bf{p^{'}(0)=3a=9}$$

and from your equation i also get $c=9$

So Is my solution is Right OR Not

Thanks

Carrotsticks · Apr 8, 2012

I'm not going to check your arithmetic, but your method is right!

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polynomial with derivative (1 Viewer)

juantheron

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Carrotsticks

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juantheron

Active Member

Carrotsticks

Retired

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