Given the function $f\left(t\right)=t^3-3t^2+2$, for which values of $t$ is the curve concave upward? (Select the correct interval.)

• A. $t<1$
• B. $1<t<2$
• C. $t<2$
• D. $t>1$

Answer: (D)

Verified step by step guidance

1. Step 1: Recall that concavity is determined by the second derivative of the function. A curve is concave upward when the second derivative is positive. Start by finding the first derivative of the function f(t) = t^3 - 3t^2 + 2.
2. Step 2: Differentiate f(t) to find f'(t). Using the power rule, f'(t) = 3t^2 - 6t.
3. Step 3: Differentiate f'(t) to find the second derivative f''(t). Again using the power rule, f''(t) = 6t - 6.
4. Step 4: Set f''(t) > 0 to determine where the curve is concave upward. Solve the inequality 6t - 6 > 0 to find the values of t.
5. Step 5: Solve the inequality 6t - 6 > 0 by isolating t. Divide both sides of the inequality by 6 to get t > 1. The curve is concave upward for t > 1.