Answer :

The gradient of the normal to the curve of h at x =3 is 1/20.

The provided is that h(x) = f(g(x))

Also provided,

g(3) = 7

g'(3) = 4

And f'(7) = -5

Taking,

h(x) = f(g(x))

Differentiating with respect to x,

h'(x) = f('g(x)).g'(x)

h'(x) is the gradient of the curve at x,

Now, putting the value of x = 3,

h'(x) = f('g(3)).g'(3)

We know, g(3) = 7 and g'(x) = 4,

h'(x) = f'(7).4

We know, f'(x) = -5

h'(x) = -5(4)

h'(x) = -20

The slope to the curve at x = 3 is -20.

We know the slope of the normal M1 and the slope at the curve M2 at the same point has their product value as -1,

So,

M1.M2 = -1

M1(-20) = -1

M1 = 1/20

So, the slope of the normal at the curve at x = 3 is 1/20.

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