YOU have forgot to check the minimum for end points

6 votes

What is the least value of the function $f(x) = 2x^{2}-8x-3$ in the interval $[0, 5]$?

- $-15$
- $7$
- $-11$
- $-3$

5 votes

Best answer

$f(x) = 2x^{2}-8x-3$

$f'(x) = 4x-8 $

For stationary point$:f'(x) = 0\implies 4x - 8 = 0 \implies x = 2$

Therefore, critical points $x = 0,2,5$

Now, $f''(x) = 4>0\:\: \textbf{(Minima)}$

For getting the minimum (or) least value , we should check all the value of critical points (stationary point and closed interval points).

For $x = 0: f(x) = -3$

For $x = 2: f(x) = 8 - 16 - 3 = -11$

For $x = 5: f(x) = 50 - 40 - 3 = 7 $

$\therefore x= 2 ,\text{minimum value of}\: f(x) = -11$

So, the correct answer is $(C).$

$f'(x) = 4x-8 $

For stationary point$:f'(x) = 0\implies 4x - 8 = 0 \implies x = 2$

Therefore, critical points $x = 0,2,5$

Now, $f''(x) = 4>0\:\: \textbf{(Minima)}$

For getting the minimum (or) least value , we should check all the value of critical points (stationary point and closed interval points).

For $x = 0: f(x) = -3$

For $x = 2: f(x) = 8 - 16 - 3 = -11$

For $x = 5: f(x) = 50 - 40 - 3 = 7 $

$\therefore x= 2 ,\text{minimum value of}\: f(x) = -11$

So, the correct answer is $(C).$

6 votes

Answer -11

f(x) = 2x^{2 }-8x -3

f'(x)=4x-8

f''(x)=4 , then it has minimal value at the point x=2

minimum value is 2⨉(2)^{2 }-8⨉(2) -3= -11