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The diagram below shows two functions $f(x)$ and $g(x)$ - Leaving Cert Mathematics - Question 4 - 2020
Question 4
The diagram below shows two functions $f(x)$ and $g(x)$. The function $f(x)$ is given by the formula $f(x) = x^3 + kx^2 + 15x + 8$, where $k \in \mathbb{Z}$, and $... show full transcript
Worked Solution & Example Answer:The diagram below shows two functions $f(x)$ and $g(x)$ - Leaving Cert Mathematics - Question 4 - 2020
Step 1
Step 2
Find the equation of $g(x)$
Answer
The turning points occur when :
3x^2 - 18x + 15 = 0$$ Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$x = \frac{18 \pm \sqrt{(-18)^2 - 4(3)(15)}}{2(3)}\ = \frac{18 \pm \sqrt{324 - 180}}{6}\ = \frac{18 \pm \sqrt{144}}{6}\ = \frac{18 \pm 12}{6}$$ Thus we find: $$x = 5\, \text{and } 1$$ Calculating $f(5)$ and $f(1)$: $$f(5) = 5^3 - 9(5^2) + 15(5) + 8 = 15\ \text{and }\ \, f(1) = 1^3 - 9(1^2) + 15(1) + 8 = -17$$ Now we find the slope of the line $g(x)$: $$m_{g(x)} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{15 - (-17)}{5 - 1} = \frac{32}{4} = 8$$ Using point-slope form of the line, we have: $$y - y_1 = m(x - x_1)\ y - 15 = 8(x - 5)\ \, y = 8x - 40 + 15\ \,y = 8x - 25$$ Thus, the equation of $g(x)$ is: $$g(x) = 8x - 25$$Step 3
Show that the graph of $g(x)$ contains the point of inflection of $f(x)$
Answer
The point of inflection occurs where . Calculate:
Setting this to zero, we find:
x = 0$$ Now we evaluate $f(0)$: $$f(0) = 0^3 - 9(0^2) + 15(0) + 8 = 8$$ Thus the point of inflection is $(0, 8)$. Now, check if $(0, 8)$ lies on $g(x)$: $$g(0) = 8(0) - 25 = -25$$ Since $(0, 8)$ does not satisfy $g(x)$, thus $g(x)$ does not contain the point of inflection.