Questions: Use the function below to answer parts (a)-(d). f(x) = (5/2) x^2 (a) Graph the function (b) Draw lines tangent to the graph at points whose x-coordinates are 0, and 1,3. (c) Find f'(x) by determining lim (f(x+h)-f(x))/h. (d) Find f'(0), f'(1), and f'(3).

Use the function below to answer parts (a)-(d).
f(x) = (5/2) x^2
(a) Graph the function
(b) Draw lines tangent to the graph at points whose x-coordinates are 0, and 1,3.
(c) Find f'(x) by determining lim (f(x+h)-f(x))/h.
(d) Find f'(0), f'(1), and f'(3).

Transcript text: Use the function below to answer parts (a)-(d). $f(x)=\frac{5}{2} x^{2}$ (a) Graph the function (b) Draw lines tangent to the graph at points whose $x$-coordinates are 0 , and 1,3 . (c) Find $f^{\prime}(x)$ by determining $\lim \frac{f(x+h)-f(x)}{h}$. (d) Find $f^{\prime}(0), f^{\prime}(1)$, and $f^{\prime}(3)$.

Solution

Solution Steps

Step 1: Graphing the function

The function $f(x) = \frac{5}{2}x^2$ is a parabola that opens upwards and is vertically stretched by a factor of 5/2. The vertex is at the origin (0,0). Graph C correctly represents this function.

Step 2: Drawing tangent lines and determining f'(x)

The formula for the derivative provided is the limit definition of a derivative. For $f(x) = \frac{5}{2}x^2$, the derivative is $f'(x) = 5x$. Graph B correctly depicts the tangent lines at x=0, x=1, and x=3. At x = 0, the tangent line is horizontal (slope = 0). At x=1, the tangent line has a positive slope, and at x=3, the tangent line has a steeper positive slope.

Step 3: Finding f'(0), f'(1), and f'(3)

Using $f'(x) = 5x$, we find the following values: $f'(0) = 5(0) = 0$ $f'(1) = 5(1) = 5$ $f'(3) = 5(3) = 15$