derivatives

[jeca86]

1) Sketch the graph of a function g for which g(0)=g'(0)=0, g'(-1)= -1, g'(1)=3, and g'(2)=1.


2)If the tangent line to y=f(x) at (4,3) passes through the point (0,2), find f(4) and f'(4).

*What does the f'(x) sign represent? That is what I don't understand mostly.

 

[tkhunny]

*What does the f'(x) sign represent? That is what I don't understand mostly.

It's the first derivative of the function f(x). Are you sure you're in the right class. You should not have missed that.

1) Sketch the graph of a function g for which g(0)=g'(0)=0, g'(-1)= -1, g'(1)=3, and g'(2)=1.

Count up the clues. I get 5. If we use a polynomial, we'll need at least a quartic.

g(x) = a*x⁴ + b*x³ + c*x² + d*x + e

Now use the clues;

g(0) = 0 = a*0⁴ + b*0³ + c*0² + d*0 + e = e, so e = 0

g(x) = a*x⁴ + b*x³ + c*x² + d*x

g'(0) = 0 g'(x) = 4*a*x³ + 3*b*x² + 2*c*x + d

g'(0) = 0 = 4*a*0³ + 3*b*0² + 2*c*0 + d = d, so d = 0

g(x) = a*x⁴ + b*x³ + c*x²
g'(x) = 4*a*x³ + 3*b*x² + 2*c*x

g'(-1) = -1 = 4*a*(-1)³ + 3*b*(-1)² + 2*c*(-1) = -4a + 3b - 2c

g'(1) = 3 = 4*a*(1)³ + 3*b*(1)² + 2*c*(1) = 4a + 3b + 2c

g'(2) = 1 = 4*a*(2)³ + 3*b*(2)² + 2*c*(2) = 32a + 12b + 4c

There are three equations to solve simultaneously.

 

[Unco]

G'day, jeca86!

2)If the tangent line to y=f(x) at (4,3) passes through the point (0,2), find f(4) and f'(4).

Recall that the derivative of a function f(x) will tell us the gradient of the tangent to the graph y=f(x) at a particular point.

We know that at the point (4,3) on the graph y=f(x), the tangent passes through (0,2). Thus the gradient of the tangent at this point, using the gradient formula (y1-y2)/(x1-x2), is:
(3 - 2)/(4 - 0) = 1/4

This means that at x=4, f'(x) = 1/4
That is, f'(4) = 1/4

You know what f(4) is because it was given to you in the question. If you can't see this, you may want to brush up on function notation. f(4) just means: "the output you get when you plug x=4 into a function", or in other words "the y-value of the point on the graph where x=4".