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7. (a) Express $-6x^2 + 24x - 25$ in the form $p(x + q)^2 + r$ - Scottish Highers Maths - Question 7 - 2019

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7.-(a)-Express-$-6x^2-+-24x---25$-in-the-form-$p(x-+-q)^2-+-r$-Scottish Highers Maths-Question 7-2019.png

7. (a) Express $-6x^2 + 24x - 25$ in the form $p(x + q)^2 + r$. (b) Given that $f(x) = -2x^3 + 12x^2 - 25x + 9$, show that $f'(x)$ is strictly decreasing for all $x... show full transcript

Worked Solution & Example Answer:7. (a) Express $-6x^2 + 24x - 25$ in the form $p(x + q)^2 + r$ - Scottish Highers Maths - Question 7 - 2019

Step 1

Show that $f'(x)$ is strictly decreasing for all $x \\in \mathbb{R}$

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Answer

Given the function: f(x)=2x3+12x225x+9f(x) = -2x^3 + 12x^2 - 25x + 9

  1. Differentiate: Calculate the first derivative: f(x)=dfdx=6x2+24x25f'(x) = \frac{df}{dx} = -6x^2 + 24x - 25

  2. Analyze the quadratic: To determine whether f(x)f'(x) is strictly decreasing, we examine the derivative f(x)f'(x), which is a quadratic function. The leading coefficient is 6<0-6 < 0, indicating that it opens downwards.

  3. Find the vertex: The vertex can be found using: x=b2a=242(6)=2x = -\frac{b}{2a} = -\frac{24}{2(-6)} = 2 We can calculate: f(2)=6(2)2+24(2)25=24+4825=1<0.f'(2) = -6(2)^2 + 24(2) - 25 = -24 + 48 - 25 = -1 < 0. So f(x)f'(x) is less than zero at the vertex.

  4. Sign of f(x)f'(x): Since the parabola opens downwards and the vertex is at a maximum of 1<0-1 < 0, we conclude that: f(x)<0f'(x) < 0 for all xinRx \\in \mathbb{R}, thus proving that f(x)f'(x) is strictly decreasing.

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