Values of [tex]x[/tex] : 15.483, 0.517
Step-by-step explanation:
Given, [tex]y=\dfrac{1}{6}x^{3}-4x^{2}+9x-1[/tex]
Differentiating both sides with respect to [tex]x[/tex], we get
[tex]\dfrac{dy}{dx}=\dfrac{1}{6}(3x^{2})-4(2x)+9(1)-0[/tex]
[tex]=\dfrac{1}{2}x^{2}-8x+9[/tex]
Since the slope is [tex]5[/tex], we write
[tex]\dfrac{dy}{dx}=5[/tex]
[tex]\Rightarrow \dfrac{1}{2}x^{2}-8x+9=5[/tex]
[tex]\Rightarrow x^{2}-16x+18=10[/tex]
[tex]\Rightarrow x^{2}-16x+8=0[/tex]
Using quadratic formula, we have
[tex]x=\dfrac{-(-16)\pm\sqrt{(-16)^{2}-4\times 1\times 8}}{2\times 1}[/tex]
[tex]=\dfrac{16\pm \sqrt{224}}{2}[/tex]
[tex]=\dfrac{16\pm 4\sqrt{14}}{2}[/tex]
[tex]=8\pm 2\sqrt{14}[/tex]
[tex]=8\pm 2\times 3.7416[/tex]
- since [tex]\sqrt{14}=3.7416...[/tex]
[tex]=8\pm 7.483[/tex]
[tex]=(8+7.483),(8-7.483)[/tex]
[tex]=15.483,0.517[/tex]