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  • If x is real, the minimum value of x2 - 8x + 17 is A. -1 B. 0 C. 1 D. 2

If x is real, the minimum value of x^2 - 8x + 17 is
A. -1
B. 0
C. 1
D. 2

Answers (1)

Let f(x)=x^2 - 8x + 17

Apply first derivative and get

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{x}^{2}-8 \mathrm{x}+17\right)}{\mathrm{dx}}\\ &\text { Apply sum rule and get }\\ &\Rightarrow f^{\prime}(x)=\frac{d\left(x^{2}\right)}{d x}-8 \frac{d(x)}{d x}+0 \end{aligned}

Apply derivative,

\Rightarrow f'(x) = 2x - 8 \\ \text{Put } f'(x) = 0 \text{ and get} \\ 2x - 8 = 0 \\ \Rightarrow 2x = 8 \\ \Rightarrow x = 4

Hence the minimum value of f(x) at x=4 is given by

\\ {f(x)= x\textsuperscript{2} - 8x + 17}\\ {f(4)=4\textsuperscript{2}-8(4)+17}\\ { \Rightarrow f(4)=16-32+17}\\ { \Rightarrow f(4)=1}\\

So, if x is real, 1 is the minimum value of x\textsuperscript{2} - 8x + 17

So the correct answer is option C.

Posted by

infoexpert22

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