Exercise 6.3 · Q11: Find $a$ when $x^4-62x^2+ax+9$ is greatest at $x=1$ on $[0,2]$

Basics you need

Solution

1. Why the derivative must vanish at $x = 1$

   Write $f(x) = x^4 - 62x^2 + ax + 9$⁠. It is a polynomial, so it is differentiable at every real number, in particular on $[0, 2]$⁠. The point $x = 1$ is an interior point of $[0, 2]$ (it is not an end point: $0 < 1 < 2$⁠). We are told that $f$ attains its maximum value on $[0, 2]$ at $x = 1$⁠. By Theorem 6(i), Section 6.4.1,

   $$f'(1) = 0.$$
   Near an interior point we can move a little to the left and a little to the right and stay inside the interval. If the tangent at $x = 1$ had a positive slope, moving right would raise $f$⁠; if negative, moving left would raise $f$⁠. Either way $f(1)$ would not be the greatest value. So the slope must be 0. At an end point this argument fails, which is why "interior" matters.
   Theorem 6 needs two things: $f$ differentiable on the closed interval, and $c$ an interior point. Both hold here. Lesson: absolute extrema

2. Differentiate and solve for $a$

   $$f'(x) = 4x^3 - 124x + a.$$

   Put $x = 1$⁠:

   $$f'(1) = 4 - 124 + a = a - 120.$$

   So $f'(1) = 0$ gives

   $$a = 120.$$
   The power rule gives $4x^3$ from $x^4$ and $-124x$ from $-62x^2$ (since $62 \times 2 = 124$⁠); $ax$ gives $a$⁠, and the constant 9 gives 0. The unknown $a$ survives differentiation, which is what lets one condition fix it.
   $\frac{d}{dx}(-62x^2) = -62 \cdot 2x = -124x$⁠. Basics: derivatives

3. The condition was only necessary: check it really is the maximum

   Theorem 6 says "if greatest at an interior point, then $f' = 0$ there"; the converse is false (a flat point can be a minimum or neither). So we check, with $a = 120$⁠:

   $$f(x) = x^4 - 62x^2 + 120x + 9,$$ $$\begin{aligned} f'(x) &= 4x^3 - 124x + 120 \\ &= 4(x^3 - 31x + 30). \end{aligned}$$

   Since $1 - 31 + 30 = 0$⁠, $x - 1$ is a factor, and dividing gives

   $$\begin{aligned} x^3 - 31x + 30 &= (x - 1)(x^2 + x - 30) \\ &= (x - 1)(x - 5)(x + 6). \end{aligned}$$ $$f'(x) = 4(x - 1)(x - 5)(x + 6).$$
   Expanding checks the factorisation: $(x - 1)(x^2 + x - 30)$ is $x^3 + x^2 - 30x - x^2 - x + 30$⁠, which is $x^3 - 31x + 30$⁠, and $x^2 + x - 30 = (x - 5)(x + 6)$ because $-5 \times 6 = -30$ and $-5 + 6 = 1$⁠.
   Factor theorem: $p(1) = 0$ means $(x - 1)$ divides $p(x)$⁠. Basics: factorising

4. Working rule on $[0, 2]$

   The critical points are $x = 1, 5, -6$⁠; only $x = 1$ lies in $[0, 2]$⁠. Evaluate $f$ there and at both end points:

   $x$	working	$f(x)$
   0 (end point)	$0 - 0 + 0 + 9$	9
   1 (critical)	$1 - 62 + 120 + 9$	68
   2 (end point)	$16 - 248 + 240 + 9$	17

   The largest of 9, 68 and 17 is 68, at $x = 1$⁠. So with $a = 120$ the maximum value of $f$ on $[0, 2]$ is indeed attained at $x = 1$⁠, as the question says.

   $f$ is continuous on the closed interval $[0, 2]$⁠, so it has a greatest value (Theorem 5), and by Theorem 6 that value occurs at a critical point inside or at an end point. Comparing the three candidates therefore finds it. At $x = 2$⁠: $2^4 = 16$⁠, $62 \times 4 = 248$⁠, $120 \times 2 = 240$⁠.
   Working Rule, Section 6.4.1: critical points in the interval, end points, compare values. Lesson: absolute extrema

5. The picture: rising, then falling

   On $[0, 2]$ the factor $x + 6$ is positive and $x - 5$ is negative. So the sign of $f'(x) = 4(x - 1)(x - 5)(x + 6)$ is the opposite of the sign of $x - 1$⁠:

   • for $0 < x < 1$⁠: $x - 1 < 0$⁠, so $f'(x) > 0$ and $f$ is increasing;
   • for $1 < x < 2$⁠: $x - 1 > 0$⁠, so $f'(x) < 0$ and $f$ is decreasing.

   Also $f''(x) = 12x^2 - 124$⁠, so $f''(1) = -112 < 0$⁠: the second derivative test agrees that $x = 1$ is a point of local maxima.

   This is a second, independent confirmation: the sign pattern $+$ then $-$ shows $f(1)$ beats every other value on $[0, 2]$⁠, not just nearby ones (Theorem 1, Section 6.3).
   Sign of a product: count the negative factors. Basics: sign of a product

6. No other value of $a$ works

   If $a \ne 120$⁠, then $f'(1) = a - 120 \ne 0$⁠, and by Theorem 6 the maximum value on $[0, 2]$ cannot be attained at the interior point $x = 1$⁠. So $a = 120$ is the only possible value, and Step 4 showed that it does work.

   "Find the value" asks for every value that satisfies the condition. This step shows no other value can work; Step 4 shows this one does. Together they settle the question completely.
   A necessary condition narrows the candidates; a check shows the candidate is genuine. Lesson: maxima and minima

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