Given f(x) a real differentiable function in x=a, the slope of the tangent line of the function in x=a is given by f'(x=a). Where f' is the first derivative of f.

We are given

$y=x-x^5$

The derivative is

$y'=1-5x^4$

(a) The slope of the tangent line at (1,0) is

$f'(1)=1-5\cdot 1^4=-4$

$f'(1)=-4$

(b) The equation of the tangent line can be found with the general formula of the line:

$y-y_o=m(x-x_o)$

Where m is the slope and the point (xo,yo) belongs to the line. We have m=-4, xo=1, yo=0, thus

$y-0=-4(x-1)$

Or, equivalently

$y+4x-4=0$

3 0

2 years ago

Gilbert walked 288 minutes. that is 4 times as many minutes as Eileen walked. how many minutes ,m, did Eileen walk?

zimovet [89]

288/4 = 72 so eileen walked for 72 minutes

3 0

2 years ago

Read 2 more answers

What is the y-intercept of a line perpendicular to y= 1/2x + 6 that passes through the point (4,-3)

Finger [1]

Answer:

(0,5)

Step-by-step explanation:

Perpendicular lines have slopes which are negative reciprocals of each other. The slope of the line y=1/2x + 6 is 1/2. The slope perpendicular to it is -2.

Using m=-2, substitute it and (4,-3) into the point slope form and simplify to find the y-intercept.

$y-y_1 = m(x-x_1)\\y--3=-2(x-4)\\y+3=-2(x-4)\\y+3=-2x+8\\y=-2x+5$

The equation is now in slope intercept form y=mx+b and b is the y-intercept. The y-intercept of y=-2x+5 is (0,5).

4 0

3 years ago