Q. Minimum value of \( (x+1)(x+2)(x+3)(x+4) \).

Q: What is the minimum value of (x+1)(x+2)(x+3)(x+4) ?

The minimum value is -1, attained at x = -\frac{5}{2} \pm \frac{\sqrt{5}}{2}.

Q: How do you find the minimum by a symmetry/substitution method?

Let u = x + \frac{5}{2}. Then the product is (u^2 - \frac{1}{4})(u^2 - \frac{9}{4}) = u^4 - \frac{5}{2}u^2 + \frac{9}{16}. Set y = u^2, minimize y^2 - \frac{5}{2}y + \frac{9}{16}; vertex at y = \frac{5}{4} gives value -1.

Q: How would you solve it using calculus?

Differentiate f(x) = (x+1)(x+2)(x+3)(x+4). Expand or use product rule to get f'(x) = 4x^3 + 30x^2 + 70x + 50. Solve f'(x)=0; critical points include x = -\frac{5}{2} \pm \frac{\sqrt{5}}{2}, where f attains the global minimum -1.

Q: Where are the zeros of the polynomial?

The zeros are x = -1, -2, -3, -4. The minimum -1 occurs between zeros, at symmetric points about x = -\frac{5}{2}.

Q: How else can the polynomial be rewritten to help minimize it?

Write (x+1)(x+4)=x^2+5x+4 and (x+2)(x+3)=x^2+5x+6. Then f=(x^2+5x)^2+10(x^2+5x)+24. Let y=x^2+5x and minimize the quadratic in y.

Answer

Let \(t = x + 2.5\). Then
\[
f(x) = (x+1)(x+2)(x+3)(x+4) = (t^2 – 0.25)(t^2 – 2.25).
\]
Put \(u = t^2 \ge 0\), so
\[
f = u^2 – 2.5u + 0.5625,
\]
a parabola with vertex at \(u = 1.25\), giving
\[
f_{\min} = 1.25^2 – 2.5\cdot 1.25 + 0.5625 = -1.
\]
Thus the minimum value is \(-1\), attained at
\[
x = -2.5 \pm \sqrt{1.25} = \frac{-5 \pm \sqrt{5}}{2}.
\]

Detailed Explanation

Problem

Find the minimum value of the function

\[ f(x) = (x+1)(x+2)(x+3)(x+4). \]

Step-by-step solution

Pair the factors to exploit symmetry.

Multiply (x+1) with (x+4) and (x+2) with (x+3):

\[ (x+1)(x+4) = x^2 + 5x + 4, \]
\[ (x+2)(x+3) = x^2 + 5x + 6. \]

So

\[ f(x) = (x^2 + 5x + 4)(x^2 + 5x + 6). \]
Introduce a new variable to simplify the product.

Let

\[ y = x^2 + 5x + 5. \]

Then

\[ x^2 + 5x + 4 = y – 1, \quad x^2 + 5x + 6 = y + 1. \]

Therefore

\[ f(x) = (y – 1)(y + 1) = y^2 – 1. \]
Minimize f by minimizing y^2.

Since y^2 is always nonnegative for real y, the expression y^2 – 1 is minimized when y^2 is as small as possible, i.e. when y^2 = 0. Thus the minimum possible value of f is

\[ \min f(x) = 0 – 1 = -1. \]
Find x-values where the minimum is attained (optional verification).

Set y = 0:

\[ x^2 + 5x + 5 = 0. \]

Solving gives

\[ x = \frac{-5 \pm \sqrt{25 – 20}}{2} = \frac{-5 \pm \sqrt{5}}{2}. \]

At these x-values, y = 0 and therefore f(x) = -1, confirming the minimum is attained.

Answer

The minimum value of (x+1)(x+2)(x+3)(x+4) is

\[ -1. \]

This minimum occurs at \[ x = \frac{-5 \pm \sqrt{5}}{2}. \]

See full solution

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FAQs

What is the minimum value of \( (x+1)(x+2)(x+3)(x+4) \)?

The minimum value is -1, attained at \(x = -\frac{5}{2} \pm \frac{\sqrt{5}}{2}\).

How do you find the minimum by a symmetry/substitution method?

Let \(u = x + \frac{5}{2}\). Then the product is \((u^2 - \frac{1}{4})(u^2 - \frac{9}{4}) = u^4 - \frac{5}{2}u^2 + \frac{9}{16}\). Set \(y = u^2\), minimize \(y^2 - \frac{5}{2}y + \frac{9}{16}\); vertex at \(y = \frac{5}{4}\) gives value -1.

How would you solve it using calculus?

Differentiate \(f(x) = (x+1)(x+2)(x+3)(x+4)\). Expand or use product rule to get \(f'(x) = 4x^3 + 30x^2 + 70x + 50\). Solve \(f'(x)=0\); critical points include \(x = -\frac{5}{2} \pm \frac{\sqrt{5}}{2}\), where \(f\) attains the global minimum -1.

Can AM-GM or other inequalities give the minimum?

AM-GM is not directly helpful because factors change sign. Symmetry/substitution or calculus is simpler. One can use completing-the-square on the centered polynomial instead.

Where are the zeros of the polynomial?

The zeros are \(x = -1, -2, -3, -4\). The minimum -1 occurs between zeros, at symmetric points about \(x = -\frac{5}{2}\).

Does the function have a maximum?

How else can the polynomial be rewritten to help minimize it?

Write \((x+1)(x+4)=x^2+5x+4\) and \((x+2)(x+3)=x^2+5x+6\). Then \(f=(x^2+5x)^2+10(x^2+5x)+24\). Let \(y=x^2+5x\) and minimize the quadratic in \(y\).

What is the graph shape and how many critical points does it have?

It's a quartic with positive leading coefficient, so it goes to \(+\infty\) at both ends. It has three critical points: two local minima (one global, value \(-1\)) and one local maximum between them.

Minimum value: (x+1)(x+2)(x+3)(x+4)
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Q. Minimum value of \( (x+1)(x+2)(x+3)(x+4) \).

Answer

Detailed Explanation

Problem

Step-by-step solution

Answer

Graph

FAQs

What is the minimum value of \( (x+1)(x+2)(x+3)(x+4) \)?

How do you find the minimum by a symmetry/substitution method?

How would you solve it using calculus?

Can AM-GM or other inequalities give the minimum?

Where are the zeros of the polynomial?

Does the function have a maximum?

How else can the polynomial be rewritten to help minimize it?

What is the graph shape and how many critical points does it have?

Similar Questions

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