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For function $f\left ( x \right )=4x^{3}-6x^{2},$ the maximum occurs in the interval $\left [ 1,2 \right ]$ when $x$ is equal to ___________

I got $x=0$ is maxima. Because, in that point $f''(x)<0$

But in answer, it is given though $f''(x)<0$ at $x=0,$ but at $x=2$ function $f(x)$ value maximum i.e. $8.$ But is it a correct procedure to get maximum.

Is maximum value and maxima are two different thing??

DDS
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Srestha
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3 Answers3

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You are told to find the maximum value of $f(x) = 4x^3 - 6x^2$ in the interval $[1,2]$. So why did you simply assert $x = 0$ when this value is clearly not in the requested interval?

This is a common mistake students make: the calculation goes flawlessly, but there is no understanding of what it means.

When searching for critical points satisfying $$\frac{df}{dx} = 12x^2 - 12x = 0,$$ we easily find $x(x-1) = 0$ or $x \in \{0, 1\}$. Therefore, these are relative extrema of $f$. By computing the second derivative, $$\frac{d^2f}{dx^2} = 24x - 12,$$ we find that at $x = 0$, $f''(0) < 0$, so the function is concave down, and at $x = 1$, $f''(1) > 0$, so the function is concave up. All this tells us is that $x = 0$ is a relative maximum, and $x = 1$ is a relative minimum.

Since the only critical point of $f$ that lies in $[1,2]$ is a relative minimum, we now have to consider the value of $f$ at the endpoints of the interval. At $x = 1$, $f(1) = -2$, and at $x = 2$, $f(2) = 8$. Moreover, $f'(x) > 0$ whenever $x > 1$. So we know that on $(1,2]$, $f$ is increasing, consequently the absolute maximum of $f$ on the interval $[1,2]$ occurs at $x = 2$ and equals $8$.

heropup
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  • but in $x=2$ point $f''(x)$ is is greater than $0$ then how this point could get a maximum value?? – Srestha Jul 29 '19 at 07:42
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    @Srestha Again, you are thinking mechanically without understanding what the computations mean. Convexity--the sign of $f''(x)$--only tells you whether the function is curving upwards, or curving downwards. It has no relevance for extrema constrained on an interval. For example, what is the maximum value of $g(x) = x$ on the interval $[1, 2]$? There aren't any critical points and the second derivative is $0$ everywhere. Stop thinking that this is a calculus problem and start thinking simpler. If I asked a student who had never seen calculus this question, they would answer it easily. – heropup Jul 29 '19 at 08:46
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Strong Hints: Identify the critical values of $f(x)$ in $[1,2]$, evaluate $f(x)$ at them, as well as $f(1)$ and $f(2)$ (as you have a closed interval)---the largest of these will be the absolute maximum of $f(x)$ over $[1,2]$.

DDS
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  • Even if the interval was an open interval, it would still be useful to find the values at the endpoints. This is because f(x) is nicely enough behaved that the limits as x approaches the endpoints are the values at the endpoints. – Jasper Jul 30 '19 at 00:15
  • If the interval is not closed, there may not be any absolute extrema. The reason we check the endpoints in addition to all critical values in the closed interval is because the Extreme Value Theorem guarantees the existence of such extrema when the function is continuous and the interval is closed. Further reasoning tells us that critical values and endpoints are the only locations for such extrema. – DDS Jul 30 '19 at 07:25
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The maximum means the highest real value that the function spits out when you take the domain to be the interval provided.

First take the derivative of $f(x)$. $f'(x)=12x^2-12x=12x(x-1)$.

Now the second derivative is: $f''(x)=24x-12$. The second derivative is clearly $>0$ in the interval $[1,2]$.

So there occurs a MINIMA at the points where the first derivative is zero in the interval $[1,2]$, ie, when $x=1$. You won't consider and worry about $x=0$ as it is not in the interval.

Now we know that $f(x)$ is a continuous function (why?), and the first derivative is always positive in the interval $[1,2]$. So it is strictly increasing in the interval.

Hence, the maximum value will be reached in the interval when $x=2$, that is the maximum value is $4(2)^3-6(2)^2= 8$.

PCeltide
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  • Also, to answer your last question, maximas and minimas are where the slope of your function changes sign, while maximums and minimums are the largest and smallest values in an interval. – PCeltide Jul 29 '19 at 07:31
  • "Maxima" is the plural of "maximum". A maximum can be local or global. So, please note that the singular or plural of the word "maximum" does not indicate which type of maximum you are dealing with. – trancelocation Jul 29 '19 at 07:37