How long, in years, will it take Andrew to settle the loan? ION 7 Determine $$ f^{\prime}(x) $$ from first principles if $$ f(x)=-2 x^{2}-1 $$. Determine: 7.2.1 $$ \quad f^{\prime}(x) $$, if it is given that $$ f(x)=-2 x^{3}+3 x^{2} $$ $$ 7.2 .2 \quad \frac{d y}{d x} $$ if $$ y=2 x+\frac{1}{\sqrt{4 x}} $$ The graph $$ y=f^{\prime}(x) $$ has a minimum turning point at $$ (1 ;-3) $$. Determine the values of $$ x $$ for which $$ f $$ is concave down.

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Let's break down the problem step by step. ### Part 1: Determine $$ f^{\prime}(x) $$ from first principles for $$ f(x) = -2x^2 - 1 $$ To find the derivative $$ f^{\prime}(x) $$ from first principles, we use the definition of the derivative: $$ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ 1. **Calculate $$ f(x+h) $$**: $$ f(x+h) = -2(x+h)^2 - 1 = -2(x^2 + 2xh + h^2) - 1 = -2x^2 - 4xh - 2h^2 - 1 $$ 2. **Calculate $$ f(x+h) - f(x) $$**: $$ f(x+h) - f(x) = (-2x^2 - 4xh - 2h^2 - 1) - (-2x^2 - 1) = -4xh - 2h^2 $$ 3. **Divide by $$ h $$**: $$ \frac{f(x+h) - f(x)}{h} = \frac{-4xh - 2h^2}{h} = -4x - 2h $$ 4. **Take the limit as $$ h \to 0 $$**: $$ f^{\prime}(x) = \lim_{h \to 0} (-4x - 2h) = -4x $$ Thus, we have: $$ f^{\prime}(x) = -4x $$ ### Part 2: Determine $$ f^{\prime}(x) $$ for $$ f(x) = -2x^3 + 3x^2 $$ We can use the power rule for differentiation here: 1. **Differentiate each term**: $$ f^{\prime}(x) = \frac{d}{dx}(-2x^3) + \frac{d}{dx}(3x^2) = -6x^2 + 6x $$ Thus, we have: $$ f^{\prime}(x) = -6x^2 + 6x $$ ### Part 3: Determine $$ \frac{dy}{dx} $$ if $$ y = 2x + \frac{1}{\sqrt{4x}} $$ 1. **Rewrite $$ y $$**: $$ y = 2x + \frac{1}{\sqrt{4x}} = 2x + \frac{1}{2\sqrt{x}} $$ 2. **Differentiate**: $$ \frac{dy}{dx} = \frac{d}{dx}(2x) + \frac{d}{dx}\left(\frac{1}{2\sqrt{x}}\right) $$ The derivative of $$ 2x $$ is $$ 2 $$, and using the power rule for $$ \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2} $$: $$ \frac{d}{dx}\left(\frac{1}{2}x^{-1/2}\right) = -\frac{1}{4}x^{-3/2} $$ Thus, we have: $$ \frac{dy}{dx} = 2 - \frac{1}{4x^{3/2}} $$ ### Part 4: Determine the values of $$ x $$ for which $$ f $$ is concave down To find where $$ f $$ is concave down, we need to find the second derivative $$ f^{\prime\prime}(x) $$ and determine where it is negative. 1. **Find $$ f^{\prime}(x) $$**: From Part 2, we have: $$ f^{\prime}(x) = -6x^2 + 6x $$ 2. **Find $$ f^{\prime\prime}(x) $$**: $$ f^{\prime\prime}(x) = \frac{d}{dx}(-6x^2 + 6x) = -12x + 6 $$ 3. **Set $$ f^{\prime\prime}(x) < 0 $$**: $$ -12x + 6 < 0 \implies -12x < -6 \implies x > \frac{1}{2} $$ Thus, $$ f $$ is concave down for: $$ x > \frac{1}{2} $$ ### Summary of Results - $$ f^{\prime}(x) = -4x $$ for $$ f(x) = -2x^2 - 1 $$ - $$ f^{\prime}(x) = -6x^2 + 6x $$ for $$ f(x) = -2x^3 + 3x^2 $$ - $$ \frac{dy}{dx} = 2 - \frac{1}{4x^{3/2}} $$ for $$ y = 2x + \frac{1}{\sqrt{4x}} $$ - $$ f $$ is concave down for $$ x > \frac{1}{2} $$ - The derivative of $$ f(x) = -2x^2 - 1 $$ is $$ f^{\prime}(x) = -4x $$. - The derivative of $$ f(x) = -2x^3 + 3x^2 $$ is $$ f^{\prime}(x) = -6x^2 + 6x $$. - The derivative of $$ y = 2x + \frac{1}{\sqrt{4x}} $$ is $$ \frac{dy}{dx} = 2 - \frac{1}{4x^{3/2}} $$. - The function $$ f $$ is concave down when $$ x > \frac{1}{2} $$.

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