HELP:how to do this differentiate q

ayumi

Im really stuck,I tried it for hours!
this not on exam papers also its ordinary maths

the first q is
a) let f(x)=x^3+2x^2-1
(x to the power of 3 and 2x to the power of 2)

i)find f ′(x),the derivative of f(x)

ii)find the slope of the tangent to the curve y=f(x) at the point where x=-1

iii)find the two values of x at which the tangents to the curve y=f(x) have a slope of 7

i really thank anyone who helps me

PunkFreud

ayumi wrote: »

Im really stuck,I tried it for hours!
this not on exam papers also its ordinary maths

the first q is
a) let f(x)=x^3+2x^2-1
(x to the power of 3 and 2x to the power of 2)

i)find f ′(x),the derivative of f(x)

ii)find the slope of the tangent to the curve y=f(x) at the point where x=-1

iii)find the two values of x at which the tangents to the curve y=f(x) have a slope of 7

i really thank anyone who helps me

i) Here you need to get the derivative. So just follow the normal rules:

f'(x) = 3x^2 + 4x

ii) Finding the derivative is just another name for finding the slope of something. So it's as simple as subbing in the value of x into the derivative (which we conveniantly found in the last part):

3(-1)^2 + 4(-1) => 3 - 4 => -1


iii)Finally, the hardest part, but actually still easy. Remember how in the last part we subbed in -1? Well this time we are trying to find that number, so we let the derivative equal to 7.

3x^2 + 4x = 7
3x^2 + 4x -7 = 0 which is a quadratic equation. So lets's solve it
(3x + 7)(x - 1) = 0
x = -7/3 ; x = 1

SOLVED

ayumi

PunkFreud wrote: »

i) Here you need to get the derivative. So just follow the normal rules:

f'(x) = 3x^2 + 4x

ii) Finding the derivative is just another name for finding the slope of something. So it's as simple as subbing in the value of x into the derivative (which we conveniantly found in the last part):

3(-1)^2 + 4(-1) => 3 - 4 => -1


iii)Finally, the hardest part, but actually still easy. Remember how in the last part we subbed in -1? Well this time we are trying to find that number, so we let the derivative equal to 7.

3x^2 + 4x = 7
3x^2 + 4x -7 = 0 which is a quadratic equation. So lets's solve it
(3x + 7)(x - 1) = 0
x = -7/3 ; x = 1

SOLVED

thank you for taking the time to do this!:)