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Question: For what real values of a and b are all extrema of the function \[f(x)={{a}^{2}}{{x}^{3}}+a{{x}^{2}}...

For what real values of a and b are all extrema of the function f(x)=a2x3+ax2−x+bf(x)={{a}^{2}}{{x}^{3}}+a{{x}^{2}}-x+b negative and the maximum is at the x0=−1{{x}_{0}}=-1 ?

Explanation

Solution

Hint : The equation should be rewritten as a function of z first. Then the first derivative should be obtained from the rewritten function. The derivative should be set equal to zero at x0=−1{{x}_{0}}=-1 for further solution of aa . And further simplification should be done to find the value of bb . So, we get the real values of aa and bb are all the extremum of the function which is negative.

Complete step by step solution:
Rewriting the given function a2x3+ax2−x+b{{a}^{2}}{{x}^{3}}+a{{x}^{2}}-x+b as a function of x we have
f(x)=a2x3+ax2−x+bf(x)={{a}^{2}}{{x}^{3}}+a{{x}^{2}}-x+b
Now, finding the first derivative of the function by differentiating it using the power rule which states that d[xn]dx=nxn−1\dfrac{d\left[ {{x}^{n}} \right] }{dx}=n{{x}^{n-1}}
Hence, we have
⇒3a2x2+2ax−d(x)dx+0\Rightarrow 3{{a}^{2}}{{x}^{2}}+2ax-\dfrac{d(x)}{dx}+0
By simplifying further, we get:
⇒3a2x2+2ax−1\Rightarrow 3{{a}^{2}}{{x}^{2}}+2ax-1
To find the maximum at x0=−1{{x}_{0}}=-1 the derivatives have to be set to zero. And then solved. Hence, we have
3a2x2+2ax−1=03{{a}^{2}}{{x}^{2}}+2ax-1=0
At x0=−1{{x}_{0}}=-1 we get
3a2(−1)2+2a(−1)−1=03{{a}^{2}}{{(-1)}^{2}}+2a(-1)-1=0
By simplifying further, we get:
3a2−2a−1=03{{a}^{2}}-2a-1=0
By using the factorisation method, we get:
3a2−3a+a−1=03{{a}^{2}}-3a+a-1=0
By simplifying further, we get:
3a(a−1)+1(a−1)=03a(a-1)+1(a-1)=0
By taking (a−1)(a-1) common from the above equation we get:
(a−1)(3a+1)=0(a-1)(3a+1)=0
From the above equation we get the value of ‘a’.
a=1, −13a=1,\,\dfrac{-1}{3}
According to the question it is given that we have to find the value of a and b in the extremum of the function should be negative that means f(x)<0f(x)<0
That means a2x3+ax2−x+b<0{{a}^{2}}{{x}^{3}}+a{{x}^{2}}-x+b<0
At x0=−1{{x}_{0}}=-1 we get:
a2(−1)3+a(−1)2−(−1)+b<0{{a}^{2}}{{(-1)}^{3}}+a{{(-1)}^{2}}-(-1)+b<0
By simplifying further, we get:
−a2+a+1+b<0-{{a}^{2}}+a+1+b<0
Now, we substitute the value of ‘a’ one by one to find the value of ‘b’.
For a=1a=1
−(1)2+(1)+1+b<0-{{(1)}^{2}}+(1)+1+b<0
By simplification we get:
1+b<01+b<0
Further solving this we get:
b<−1b<-1
That means ⇒b∈(−∞,−1)\Rightarrow b\in (-\infty ,-1)
Similarly, for a=−13a=\dfrac{-1}{3}
−(−13)2+(−13)+1+b<0-{{\left( \dfrac{-1}{3} \right)}^{2}}+\left( \dfrac{-1}{3} \right)+1+b<0
By simplification we get:
−(19)+(−13)+1+b<0-\left( \dfrac{1}{9} \right)+\left( \dfrac{-1}{3} \right)+1+b<0
By taking LCM we get:
(−1−39)+1+b<0\left( \dfrac{-1-3}{9} \right)+1+b<0
By further solving we get:
−4+99+b<0\dfrac{-4+9}{9}+b<0
Now, finally we get the value of “b”.
b<−59b<\dfrac{-5}{9}
That means ⇒b∈(−∞,−59)\Rightarrow b\in \left( -\infty ,\dfrac{-5}{9} \right)
Therefore, for a=1a=1 there is b∈(−∞,−1)b\in (-\infty ,-1)
For a=−13a=\dfrac{-1}{3} there is b∈(−∞,−59)b\in \left( -\infty ,\dfrac{-5}{9} \right) when the function is negative.
So, the correct answer is “Option B”.

Note : Finding the first derivative of the function by differentiating it using the power rule is important. To find the maximum value the derivative should be set at zero and then solved. Students have to read the question carefully to find the values of a and b at which the function is negative that means function is less than zero. If the question is asked to find the values at which function is positive then only function greater than zero. Questions may also be asked in exams related to local maxima and local minima. In that case we need to take the derivative twice and if the second derivative is positive then it is local minima otherwise it is local maxima. So, you have to remember the conditions to solve this type of problem.