MCQ
If x is real, the minimum value of $x^2-8 x+17 \text { is }$
- A-1
- B$0$
- ✓1
- D2
✓
Answer
Correct option: C.
1
(C) 1
Explanation: Let,
$f(x)=x^2-8 x+17$
On differentiating with respect to x, we get
f'(x) = 2x - 8
So, f'(x) = 0
$\Rightarrow 2 x-8=0$
$\Rightarrow \quad 2 x=8$
x = 4
Now, Again on differentiating w.r.t. x, we get
$f^{\prime \prime}(x)=2>0, \forall x$
So, x = 4 is the point of local minima.
Minimum value of f(x) at x = 4
f(4) = 4 x 4 - 8 x 4 + 17 = 1
Explanation: Let,
$f(x)=x^2-8 x+17$
On differentiating with respect to x, we get
f'(x) = 2x - 8
So, f'(x) = 0
$\Rightarrow 2 x-8=0$
$\Rightarrow \quad 2 x=8$
x = 4
Now, Again on differentiating w.r.t. x, we get
$f^{\prime \prime}(x)=2>0, \forall x$
So, x = 4 is the point of local minima.
Minimum value of f(x) at x = 4
f(4) = 4 x 4 - 8 x 4 + 17 = 1
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