Differentiating a Polynomial Function

1. Find the derivative of \(f( x) =2x^{4}-6x^{2}+2x-3\).
2. What is \(f^\prime (2) ?\)
3. Find the slope of the tangent line to the graph of \(f\) at the point \( ( 1,-5)\).
4. Find an equation of the tangent line to the graph of \(f\) at the point \(( 1,-5)\).
5. Use graphing technology to graph \(f\) and the tangent line to the graph of \(f\) at the point \((1, -5)\) on the same screen.

Solution (a) \[ \begin{eqnarray*} f^\prime ( x) &=&\dfrac{d}{dx}( 2x^{4}-6x^{2}+2x-3) {=} \dfrac{d}{dx}( 2x^{4}) -\dfrac{d}{dx}( 6x^{2}) +\dfrac{d}{dx} ( 2x) -\dfrac{d}{dx}3 \\[-6.4pt] &&\hspace{4.85pc}\underset{{\color{#0066A7}{\text{Sum & Difference Rules}}}}{\color{#0066A7}{\uparrow }}\\[-6.4pt] &{=}& 2\cdot \dfrac{d}{dx}x^{4}-6\cdot \dfrac{d}{dx}x^{2}+2\cdot \dfrac{d}{dx}x-0 \\[-6.4pt] &&\hspace{-3.15pc}\underset{{\color{#0066A7}{\text{Constant Multiple Rule}}}}{\color{#0066A7}{\uparrow }}\\ &{=}& 2\cdot 4x^{3}-6\cdot 2x+2\cdot 1 {=} 8x^{3}-12x+2\\[-6.4pt] &&\hspace{-3.7pc}\hspace{1pc}\underset{\color{#0066A7}{\text{Simple Power Rule}}}{\color{#0066A7}{\uparrow }}\hspace{4.4pc}\underset{{\color{#0066A7}{\text{Simplify}}}}{\color{#0066A7}{\uparrow }} \end{eqnarray*} \]

(b) \(f^\prime (2) =8( 2)^{3} -12(2) +2=64-24+2=42\).

(c) The slope of the tangent line at the point (1, −5) equals \(f′(1)\). \[ f^\prime (1) =8( 1)^{3} -12(1) +2=8-12+2=-2 \]

(d) We use the point-slope form of an equation of a line to find an equation of the tangent line at \((1,-5)\). \[ \begin{eqnarray*} y-( -5) &=&-2( x-1) \\ y &=&-2( x-1) -5=-2x+2-5=-2x-3 \end{eqnarray*} \]

The line \(y=-2x-3\) is tangent to the graph of \(f( x) =2x^{4}-6x^{2}+2x-3\) at the point \((1, -5)\).

(e) The graphs of \(f\) and the tangent line to \(f\) at \((1, -5)\) are shown in Figure 22.