Question

Find the absolute minimum value of the function \(\rm y=3x^{2}-4\) in the interval \(\left [ -1,5 \right ]\)
1. -4
2. 4
3. 3
4. 0
5. 1

Answer 1

Correct Answer - Option 1 : -4

Concept:

Following steps to finding the absolute minimum.

Differentiate the function and equate it with zero to get the critical point 

Find the value of the function at critical point c

Find the function value at end of the interval [a, b]

Absolute minimum value is {min f(a), f(b) and f(c)}

Calculation:

\(\rm ⇒ y=3x^{2}-4\), x ∈ [-1, 5]

Differentiation with respect to x

⇒  y' = 6x

For critical points, y' = 0

⇒ x = 0

Now,

⇒ y(0) = -4

⇒ y(-1) = -1

\(\rm ⇒ y(-2)=3(2)^{2}-4\)

\(\rm⇒ y(-2)=8\)

\(\rm Min\left \{ -4,-1,8 \right \}=-4\)

\(\therefore\) Minimum value of function in given interval is -4

Hence, option 1 is correct