For which of the following values of real number $t$, the equation $x^4-tx+\dfrac 1t = 0$ has no root on the interval $[1,2]$. I think we might check for which $t$ $y=tx-\dfrac 1t $ have solution on [1,16].
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Do you mean $f(x)=x^4-tx+\frac{1}{t}$? – Ty. Jun 14 '20 at 20:38
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No, I mean f(x)=tx-1/t on [1;16] – piteer Jun 14 '20 at 21:08
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How to solve this? – piteer Jun 15 '20 at 07:48