Plz help line tangent

Question

Suppose the line tangent to the graph of f at x=3 is y=4x+2 and suppose y=5x-5 is the line tangent to the graph of g at x=3. Find an equation of the line tangent to the following curves at x=3.

a. y=f(x)+g(x) at x=3

y=?

b. y=f(x)-4g(x) at x=3

y=?

c. 4f(x) at x=3

y=?

Answer 1

If the line tangent to the graph of f(x) at x=3 is y=4x+2, then f(3)= 4*3 + 2 = 14 and f’(3) = 4

similarly if the line tangent to the graph of g(x) at x=3 is y=5x-5, then g(3)= 5*3 - 5 = 10 and g’(3) = 5

then we use the fact that derivative of a sum equals the sum of derivatives to solve the problems

i) h(x) =f(x)+g(x) at x=3

h(3) =f(3)+g(3) = 14 + 10 = 24

h’(3) =f’(3)+g’(3) = 4 + 5 = 9

That tells us that the tangent line passes through the point (3 , 24) and its slope equals 9 therefore the equation of the tangent line is

y = 9x + b, where b is a constant.

also at x=3 y=24, therefore 24=27+b, and b = -3

hence the equation of the tangent line is

y = 9x - 3

The other two problems can be solved using the same method

ii) h(x) =f(3)-4g(x) at x=3

h(3) =f(3)-4g(3) = 14 - 40 = -26

h’(3) =f’(3)-4g’(3) = 4 - 20 = -16

That tells us that the tangent line passes through the point (3 , -26) and its slope equals -16 therefore the equation of the tangent line is

y = -16x + b, where b is a constant.

also at x=3 y= -26, therefore -26 = -16*3 + b, and b = 22

hence the equation of the tangent line is

y = -16x + 22

iii) h(x) = 4f(x) at x=3

h(3) =4f(3) = 14 * 4 = 56

h’(3) =4f’(3) = 4 * 4 = 16

That tells us that the tangent line passes through the point (3 , 56) and its slope equals 16 therefore the equation of the tangent line is

y = 16x + b, where b is a constant.

also at x=3 y= 56, therefore 56 = 16*3 + b, and b = 8

hence the equation of the tangent line is

y = 16x + 8

Answer 2

The tangent line of a graph is the derivative of the parent function of that graph. Therefore, for the tangent line of f,

f'(x) = 4x + 2

so we must integrate to determine the value of the function f(x). By integration, we should get

f(x) = 2x² + 2x + c

We can calculate our value of c by plugging in the point (x, 4x+2). From this, we get the point (3, 14), which we plug into f(x) to generate a value of c = -10. Therefore

f(x) = 2x² +2x - 10

We follow the same steps for the tangent line of g.

g'(x) = 5x - 5

g(x) = 5/2x² - 5x + c

(x, 5x-5) --> (3, 10)

g(x) = 5/2x² - 5x + 5/2

From there, plug the parent functions into your problems, simplify, and take the derivative to determine the tangent line. For example,

c. 4f(x) @ x=3

4f(x) = 4(2x²+2x-10) = 8x² + 8x - 40

Derivative: y = 16x + 8

I hope this helps.