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Why does differentiation work like it does.

Differentiation is about the tangent at a point, from GCSE we have a formula to work out the gradient between two points, in other words, the gradient of the tangent. This means that as the two points get...

LF
Answered by Lucas F. Maths tutor
3011 Views

(a) Express (1+4*sqrt(7))/(5+2*sqrt(7)) in the form a+b*sqrt(7), where a and b are integers. (b) Then solve the equation x*(9*sqrt(5)-2*sqrt(45))=sqrt(80).

(a)   We can ‘get rid of’ a square root in the denominator simply by multiplying by 1 (value of the fraction stays unchanged) in a suitable form. We will take advantage of this formula: (a+b)(a-b)=a^2...

KB
Answered by Kristina B. Maths tutor
7730 Views

Find the equation of the tangent to the curve y=3x^2-7x+5 at the point (2, 3) .

The starting point for a question like this is to differentiate the function - in this case the curve y=3x2 -7x+5 . We calculate that dy/dx=6x-7 . The question tells us that we are interested i...

MS
Answered by Matthew S. Maths tutor
7325 Views

Find the equation of the line perpendicular to the line y= 3x + 5 that passes through the point (-1,4)

Logically work through the problem: 1) Plot all the information that is available so you can visualise the problem better (always encouraged for graphical questions) 2) Understand that the gradient of the...

CC
Answered by Chagall C. Maths tutor
3348 Views

Given the function f(x)=λx^3 + 9, for λ other than zero, find the inflection point of the graph in terms of λ. How does the slope of the line tangent to the inflection point changes as λ varies from 0 to 1?

f'(x) = 3λx^2f''(x) = 6λxFor the inflection point (x0,y0), it is true that f''(x0)=0 so 6λxo=0 => x0= 0 (since λ cannot be zero)Therefore, the inf...

CD
Answered by Claire D. Maths tutor
2292 Views

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