how can i show that the graph of the function f(x) = (x^5) + 3(x^5) + 5x does not have a tangent line with?

Question

with a slope of 3 ??

thnx for your help

Answer 1

Take the derivative and show that it's nowhere equal to 3.


Doug

Answer 2

Differentiate f(x), for a function ax^n, the derivative is nax^(n-1)
and differentiating term by term is valid.
so,
f(x) = 4x^5 +5x
f'(x) = 20x^4 + 5

Note that for any real x, x^2 >= 0 ,(i.e. greater than or equal to)
so x^4 >= 0 ,(as this is [x^2]^2)
then f'(x) = 20x^4 + 5 >= 5
because the 5 is independant of x and x=0 is the only x that will give you the least value of f(x). Clearly f'(x) > 3
Note that, if you're new to calculus, f'(x) is the equation of the tangent of the function. Also the f(x) you've given can be simplified further as there are like terms both in x^5
i.e f(x) = 4x^5 + 5x

Hope this helps!!

Answer 3

As pointed out before, you need to find the derivative of the function first. This derivative represents the slope of the curve as a function of x.

f(x) = 4(x^5) + 5x

f'(x) = 20 x^4 + 5

Find the value x that makes f'(x) = 3

3 = 20 x^4 +5

20 x^4 = -2

x^4 = -1/10

note that we are pursing the fourth root of a negative number...thus, there is no real value of x for which the slope of the curve will be equal to 3.

QED. Good luck.

Answer 4

you have

f(x) = 4(x^5) + 5x
df/dx = 20 x^4 + 5
miinimimum value of df/dx (same as slope of tangent ) = 5
as 20x^4 >=0
so tangent cannot have slope value < 5
so slope cannot be 3

Answer 5

you need
f(x)=(x^5) + 3(x^5) + 5x=4x^5 +5x

f'(x)=20x^4 +5=3
then 20x^4= -2
but this is impossible, since x^4 is always >=0.

Answer 6

Please recheck your question. The first and second terms seem to have the same power of x.

My bet would be to differentiate f(x) w.r.t. x. Equate the resultant to 3 and prove that the resultant equation in x does not have a real solution.