Q. 1TF - Calculus

Question

Taking the limit: We have seen that if f is a smooth function, then  $f^{\text{'}}\left(c\right)\approx \frac{f(c+h)-f\left(c\right)}{h}$This approximation should get better as h gets closer to zero. In fact, in the next section we will define the derivative in terms of such a limit.

$f^{\text{'}}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h}$.

Use the limit just defined to calculate the exact slope of the tangent line to $f\left(x\right)=x^2$ at $x=4 .$

Step-by-Step Solution

Answer

The exact slope of the tangent is 8.

Step 1.Given Information

Given equation is $f\left(x\right)=x^2$ at $x=4 .$

Step 2.Find the derivative

Formula for derivative $f^{\text{'}}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h}$

Here $$\begin{gathered} x\to c;x\to 4; \\ c\to 4 \end{gathered}$$

$$\begin{gathered} f^{\text{'}}\left(4\right)=\underset{h\to 0}{\lim }\frac{f(4+h)-f\left(4\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\left(4+h\right)}^2-{\left(4\right)}^2}{h} \\ =\underset{h\to 0}{\lim }\frac{{\cancel{\left(4\right)}}^2+h^2+8h-\cancel{{\left(4\right)}^2}}{h} \\ =\underset{h\to 0}{\lim }\frac{h^2+8h}{h} \\ =\underset{h\to 0}{\lim }\frac{\cancel{h}(h+8)}{\cancel{h}} \\ =(0+8) \\ =8 . \end{gathered}$$

Step 3.The solution

The exact slope of the tangent is 8.Since the derivative of the equation at point gives the slope.