let $f(x)=5x^{2}+12x - 2$. using the definition of derivative, $f(x)=lim_{h\rightarrow0}\frac{f(x +…

let $f(x)=5x^{2}+12x - 2$. using the definition of derivative, $f(x)=lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$, enter the expression needed to find the derivative function. $f(x)=lim_{h\rightarrow0}square$ after evaluating this limit, we see that $f(x)=\frac{df}{dx}=square$ finally, the equation of the tangent line to $f(x)$ where $x = 3$ is $square$

let $f(x)=5x^{2}+12x - 2$. using the definition of derivative, $f(x)=lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$, enter the expression needed to find the derivative function. $f(x)=lim_{h\rightarrow0}square$ after evaluating this limit, we see that $f(x)=\frac{df}{dx}=square$ finally, the equation of the tangent line to $f(x)$ where $x = 3$ is $square$

Answer

Explanation:

Step1: Find $f(x + h)$

[ \begin{align*} f(x+h)&=5(x + h)^2+12(x + h)-2\ &=5(x^{2}+2xh+h^{2})+12x + 12h-2\ &=5x^{2}+10xh+5h^{2}+12x + 12h-2 \end{align*} ]

Step2: Substitute $f(x + h)$ and $f(x)$ into the derivative - formula

[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}\frac{(5x^{2}+10xh+5h^{2}+12x + 12h-2)-(5x^{2}+12x - 2)}{h}\ &=\lim_{h\rightarrow0}\frac{5x^{2}+10xh+5h^{2}+12x + 12h-2 - 5x^{2}-12x + 2}{h}\ &=\lim_{h\rightarrow0}\frac{10xh+5h^{2}+12h}{h} \end{align*} ]

Step3: Simplify the expression

[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{h(10x + 5h+12)}{h}\ &=\lim_{h\rightarrow0}(10x + 5h+12) \end{align*} ]

Step4: Evaluate the limit

[ f^{\prime}(x)=10x + 12 ]

Step5: Find the slope of the tangent - line at $x = 3$

When $x = 3$, $m=f^{\prime}(3)=10\times3+12=42$

Step6: Find $f(3)$

[ \begin{align*} f(3)&=5\times3^{2}+12\times3-2\ &=5\times9+36 - 2\ &=45+36 - 2\ &=79 \end{align*} ]

Step7: Use the point - slope form $y - y_{1}=m(x - x_{1})$ to find the equation of the tangent line

Using the point $(3,79)$ and $m = 42$, we have $y-79=42(x - 3)$ [ \begin{align*} y-79&=42x-126\ y&=42x-47 \end{align*} ]

Answer:

$f^{\prime}(x)=\lim_{h\rightarrow0}\frac{10xh+5h^{2}+12h}{h}$ $\frac{df}{dx}=10x + 12$ $y = 42x-47$