Q. graph of \( f(x) = x^{2} – 7 \lvert x – 2 \rvert \cdot \lvert x – 3 \rvert + 5. \)

Q: How do I rewrite f as a piecewise function?

Split at x=2 and x=3: x\le2 or x\ge3:\; f(x)=-6x^2+35x-37. 2\le x\le3:\; f(x)=8x^2-35x+47.

Q: Is f continuous and differentiable at x=2 and x=3?

f is continuous everywhere. It is not differentiable at x=2 and x=3 (kinks). Left/right derivatives: at x=2 left 11, right -3; at x=3 left 13, right -1.

Q: Where are the x-intercepts (roots) of f?

Roots come from the outer quadratic -6x^2+35x-37=0: x=\dfrac{35\mp\sqrt{337}}{12}, approximately 1.3869 and 4.4465. The middle quadratic has no real roots.

Q: What is the end behavior of f?

As x \to \pm\infty, the outer quadratic -6x^2 + 35x - 37 dominates, so f(x) \to -\infty.

Q: Is f symmetric (even/odd)?

No. f(-x)\neq f(x) and f(-x)\neq -f(x); there is no even or odd symmetry.

Answer

Split at x=2 and x=3. Note A = |x-2||x-3| = x^2-5x+6 for x≤2 or x≥3, and A = −(x^2-5x+6) for 2≤x≤3. Hence

\[
f(x)=x^2-7|x-2||x-3|+5
=
\begin{cases}
-6x^2+35x-37, & x\le2\ \text{or}\ x\ge3,\\[4pt]
8x^2-35x+47, & 2\le x\le3.
\end{cases}
\]

Values at the joins: f(2)=9, f(3)=14 (continuous). The middle piece (2≤x≤3) is upward-opening with vertex at

\[
x_v=\frac{35}{16},\qquad f(x_v)=\frac{279}{32}.
\]

Thus the graph consists of the downward parabola -6x^2+35x-37 on (−∞,2] and [3,∞) joined to the upward parabola 8x^2-35x+47 on [2,3]; the global minimum is at (35/16,\,279/32).

Detailed Explanation

To graph the given function, we must remove the absolute value signs by analyzing the intervals defined by the critical points \(x = 2\) and \(x = 3\).

\[
f(x) = x^2 – 7|x – 2| \cdot |x – 3| + 5
\]

Case 1 is when \(x < 2\). Both expressions inside the absolute values are negative, making \(|x – 2| = -(x – 2)\) and \(|x – 3| = -(x – 3)\). The product of these two negatives is positive, so the function becomes \(f(x) = x^2 – 7(x – 2)(x – 3) + 5\). Expanding the product yields \(f(x) = x^2 – 7(x^2 – 5x + 6) + 5 = -6x^2 + 35x – 37\). This represents a downward-opening parabola for this interval.

Case 2 is when \(2 \le x \le 3\). Here, \(x – 2 \ge 0\) and \(x – 3 \le 0\), making \(|x – 2| = x – 2\) and \(|x – 3| = -(x – 3)\). The function becomes \(f(x) = x^2 + 7(x – 2)(x – 3) + 5\). Expanding this yields \(f(x) = x^2 + 7(x^2 – 5x + 6) + 5 = 8x^2 – 35x + 47\). This represents an upward-opening parabola for this interval.

Case 3 is when \(x > 3\). Both expressions are positive, making \(|x – 2| = x – 2\) and \(|x – 3| = x – 3\). The function evaluates identically to Case 1, giving \(f(x) = -6x^2 + 35x – 37\).

To map the graph accurately, we evaluate the function at the boundary points to ensure continuity. At \(x = 2\), \(f(2) = (2)^2 – 0 + 5 = 9\). At \(x = 3\), \(f(3) = (3)^2 – 0 + 5 = 14\).

We then find the vertex of the middle segment. For the parabola \(y = 8x^2 – 35x + 47\), the x-coordinate of the vertex is \(x = \frac{35}{16} = 2.1875\). The corresponding y-coordinate is \(f(2.1875) \approx 8.72\).

Thus, the graph connects an upward-curving arc between \( (2, 9) \) and \( (3, 14) \) reaching a local minimum at exactly \( (2.1875, 8.71875) \). This middle section is flanked by steep downward-opening parabolic curves extending to negative infinity on both the left and right sides.

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Algebra FAQs

What is the domain of \(f(x)=x^2-7|x-2||x-3|+5\)?.

Domain is all real numbers, since polynomials and absolute values are defined for every real \(x\).

How do I rewrite \(f\) as a piecewise function?

Split at x=2 and x=3: \(x\le2\) or \(x\ge3:\; f(x)=-6x^2+35x-37\). \(2\le x\le3:\; f(x)=8x^2-35x+47\).

Is \(f\) continuous and differentiable at \(x=2\) and \(x=3\)?

\(f\) is continuous everywhere. It is not differentiable at \(x=2\) and \(x=3\) (kinks). Left/right derivatives: at \(x=2\) left \(11\), right \(-3\); at \(x=3\) left \(13\), right \(-1\).

Where are the x-intercepts (roots) of \(f\)?

Roots come from the outer quadratic \(-6x^2+35x-37=0\): \(x=\dfrac{35\mp\sqrt{337}}{12}\), approximately \(1.3869\) and \(4.4465\). The middle quadratic has no real roots.

What is the y-intercept \(f(0)\)?

\(f(0)=-37\).

Where are local extrema?

What is the end behavior of \(f\)?

As \(x \to \pm\infty\), the outer quadratic \(-6x^2 + 35x - 37\) dominates, so \(f(x) \to -\infty\).

Is \(f\) symmetric (even/odd)?

No. \(f(-x)\neq f(x)\) and \(f(-x)\neq -f(x)\); there is no even or odd symmetry.

How should I sketch the graph quickly?

Steps: (1) Split into the three piecewise quadratics. (2) Plot roots \(1.387\) and \(4.447\), y-intercept \(-37\). (3) Mark cusps at \(x=2\) \(9\) and \(x=3\) \(14\) and middle minimum at \(x=\frac{35}{16}\). (4) Connect with the corresponding parabolic shapes and show tails going to \(-\infty\)..

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