Question

Let \(g:\mathbb{R}\to\mathbb{R}\) be a function defined by

• A. \(g\) has a local minimum at \(x=1\)
• B. \(g\) has a point of inflection at \(x=0\)
• C. \(x=0,\ x=1,\ x=2\) are critical points of \(g\)
• D. \(g''(x)=0\) has at least two distinct solutions, where \(g''\) denotes the second order derivative of \(g\)

Hint:

To test local maxima and minima, check the sign change of the first derivative. If \(g'(x)\) changes from positive to negative, the point is a local maximum; if it changes from negative to positive, the point is a local minimum.

Answer 1

The Correct Option is A

Step 1: Differentiate the integral.
With $g(x)=\int_0^{x^3}(t^2-9t+8)\,dt$, the chain rule gives \[ g'(x)=3x^2\,(x^6-9x^3+8)=3x^2(x^3-1)(x^3-8). \]

Step 2: List the critical points.
Setting $g'(x)=0$ gives $x=0,\,1,\,2$. So option (C) is fine and these are genuine critical points.

Step 3: Test $x=1$ by sign change.
Just below $1$ both $(x^3-1)$ and $(x^3-8)$ are negative, so $g'>0$. Between $1$ and $2$ the first is positive and the second negative, so $g'<0$. The slope turns from $+$ to $-$, which is a local maximum, not a minimum. So statement (A) fails.

Step 4: Confirm the others hold.
$g''(x)=3x(8x^6-45x^3+16)$. The lone factor $x$ makes the sign flip at $0$, so $x=0$ is an inflection point and (B) is true. Putting $y=x^3$ in $8x^6-45x^3+16=0$ gives $8y^2-45y+16=0$, whose discriminant $2025-512>0$ yields two distinct roots, so $g''=0$ has several solutions and (D) is true.

Step 5: Pick the false one.
The statement that does not hold is option (A).
\[ \boxed{(A)} \]