IB Maths AI HL The Second Derivative and Concavity Questions

1. Differentiate the function once to find $f^{\prime }(x)$: $f^{\prime }(x)=8x^3-9x^2+5$.

2. Differentiate the first derivative to find the second derivative $f^{″}(x)$: $f^{″}(x)=24x^2-18x$.

3. Substitute $x=1$ into the second derivative equation.

4. Evaluate: $f^{″}(1)=24(1{)}^2-18(1)=24-18$.

5. The exact value is $6$.

1. Find the first derivative: $\frac{dy}{dx}=3x^2-12x+9$.

2. Find the second derivative: $\frac{d^{2}y}{d{x}^2}=6x-12$.

3. Set the second derivative to zero to locate potential points of inflexion: $6x-12=0⟹6x=12⟹x=2$.

4. Substitute $x=2$ back into the original curve equation to find the $y$-coordinate.

5. Evaluate: $y=(2{)}^3-6(2{)}^2+9(2)+2=8-24+18+2=4$.

6. The exact coordinates of the point of inflexion are $(2,4)$.

1. Find the first derivative: $f^{\prime }(x)=-6x^2+12x$.

2. Set $f^{\prime }(x)=0$ to find stationary points: $-6x(x-2)=0⟹x=0$ and $x=2$.

3. Find the second derivative: $f^{″}(x)=-12x+12$.

4. Test $x=0$ using the second derivative: $f^{″}(0)=-12(0)+12=12$. Since $12>0$, the curve is concave up, so $x=0$ is a local minimum.

5. Test $x=2$ using the second derivative: $f^{″}(2)=-12(2)+12=-12$. Since $-12<0$, the curve is concave down, so $x=2$ is a local maximum.